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www.jeebooks.in, Skills in, I', , t, , I, , i, , J, , 'I, , Mathematics for, , JEE MAIN &, ADVANCED, , Lr, , o, , OJ, , 0, , With Sessionwise Theory & Exercises, , Practice all Objective Questions from, I, this book on your mobile for FREE, W Detai/ed Instructions inside---------------->, , o C o, , www.jeebooks.in, Amit M. Agarwal, , v, , arihant

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www.jeebooks.in, I, , Skills in Mathematics for, , JEE MAIN & ADVANCED, , arihant, ARIHANT PRAKASHAN (Series), MEERUT, All Rights Reserved, , *, , ©AUTHOR, No part of this publication may be re-produced, stored in a retrieval system or by any means, lectronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the, publisher. Arihant has obtained all the information in this book from the sources believed to be reliable, and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the, absolute accuracy of any information published, and the damages or loss suffered thereupon., All disputes subject to Meerut (UP) jurisdiction only., , ADMINISTRATIVE & PRODUCTION OFFICES, Regd. Office, ‘Ramchhaya 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002, Tele: Oil- 47630600,43518550; Fax: Oil- 23280316, Head Office, Kalindi, TP Nagar, Meerut (UP) - 250002, , Tel: 0121-2401479,2512970,4004199; Fax: 0121-2401648, , SALES & SUPPORT OFFICES, Agra, Ahmedabad, Bengaluru, Bhubaneswar, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur,, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune, , ISBN : 978-93-12146-92-7, Price : 7 255.00, Printed & Bound by Arihant Publications (I) Ltd. (Press Unit), , www.jeebooks.in, For further information about the books from Arihant,, log on to www.arihantbooks.com or email to [email protected]

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www.jeebooks.in, t, , Skills in Mathematics for, , JEE MAIN & ADVANCED, , PREFACE, “YOU CANDO ANYTHING IF YOU SET YOUR MIND TO IT, I TEACH TRIGONOMETRY, TO JEE ASPIRANTS BUT BELIEVE THE MOST IMPORTANT FORMULA IS, COURAGE + DREAMS = SUCCESS”, It is a matter of great pride and honour for me to have received such an overwhelming, response to the previous editions of this book from the readers. In a way, this has inspired, me to revise this book thoroughly as per the changed pattern of JEE Main & Advanced. I, have tried to make the contents more relevant as per the needs of students, many topics, have been re-written, a lot of new problems of new types have been added in etc. All, possible efforts are made to remove all the printing errors that had crept in previous, editions. The book is now in such a shape that the students would feel at ease while going, through the problems, which will in turn clear their concepts too., , A Summary of changes that have been made in Revised & Enlarged Edition, •, , Theory has been completely updated so as to accommodate all the changes made in JEE, Syllabus & Pattern in recent years., , •, , The most important point about this new edition is, now the whole text matter of each, chapter has been divided into small sessions with exercise in each session. In this way the, reader will be able to go through the whole chapter in a systematic way., , •, , Just after completion of theory, Solved Examples of all JEE types have been given, providing, the students a complete understanding of all the formats of JEE questions & the level of, difficulty of questions generally asked in JEE., , •, , Along with exercises given with each session, a complete cumulative exercises have been, given at the end of each chapter so as to give the students complete practice for JEE along, with the assessment of knowledge that they have gained with the study of the chapter., , •, , Previous Years questions asked in JEE Main &Adv, IIT-JEE & AIEEE have been covered in, all the chapters., , However I have made the best efforts and put my all teaching experience in revising this, book. Still I am looking forward to get the valuable suggestions and criticism from my, own fraternity i.e. the fraternity of JEE teachers., , I would also like to motivate the students to send their suggestions or the changes that, they want to be incorporated in this book. All the suggestions given by you all will be, kept in prime focus at the time of next revision of the book., , www.jeebooks.in, Amit M. Agarwal

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www.jeebooks.in, t jr Skills in Mathematics for, , Ox JEE MAIN & ADVANCED, , CONTENTS, 1. TRIGONOMETRIC FUNCTIONS, AND IDENTITIES, LEARNING PART, Session 1, • Measurement of Angles, , Session 7, , Session 2, • Definition of Trigonometric Functions, , Session 8, , Session 3, • Application of Basic Trigonometry on, Eliminating Variables or Parameters, and Geometry, , 1-128, , • Sum of Sines/Cosines in Terms of, Products, • Trigonometric Ratios of Multiples of, an Angle, Session 9, , • Trigonometric Ratios of sub multiple, of an Angle, , Session 4, • Signs and Graph of Trigonometric, Functions, , Session 10, • Trigonometric Ratios of the Sum of, Three or More Angles, , Session 5, , Session 11, • Maximum and Minimum Values of, Trigonometrical Functions, , • Trigonometric Ratios of any Angle, Session 6, , • Trigonometric Ratios of Compound, Angles, , PRACTICE PART, • JEE Type Examples, • Chapter Exercises, , 2. TRIGONOMETRIC EQUATIONS, AND INEQUATIONS, LEARNING PART, Session 1, • Trigonometric Equations, • Principal Solution, • General Solution, Session 2, • Equation of the Form, a cos q + b sin q = c, • Some Particular Equations, , 129-196, , Session 3, , • Solution of Simultaneous, Trigonometric Equations, • Problems Based on Extreme, Values of sin x and cos x, Session 4, , • Trigonometric Inequality, , www.jeebooks.in, PRACTICE PART, • JEE Type Examples, • Chapter Exercises

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www.jeebooks.in, J, , Skills in Mathematics for, , JEE MAIN & ADVANCED, , 3. PROPERTIES AND SOLUTIONS, OF TRIANGLES, LEARNING PART, , Session 5, , Session 1, , •, , •, , Basic Relation between the Sides and, Angles of Triangle, , Session 2, • Auxiliary Formulae, , Regular Polygons and Radii of the, Inscribed and Circumscribing Circle, a Regular Polygon, , Session 6, , •, , Quadrilaterals and, Cyclic Quadrilaterals, , Session 3, • Circles Connected with Triangle, , Session 7, , Session 4, , Session 8, , •, /, , •, , Orthocentre and its Distance from, the Angular Points of a Triangle and, Pedal Triangle, Centroid of Triangle, , •, •, , Solution of Triangles, Height and Distance, , PRACTICE PART, • JEE Type Examples, • Chapter Exercises, , 4. INVERSE TRIGONOMETRIC, FUNCTIONS, LEARNING PART, Session 1, • Inverse Trigonometric Functions, • Inverse Function, • Domain and Range of Inverse, Trigonometric Functions, Session 2, , •, , Property I of Inverse Trigonometric, Functions, , Session 3, • Property II of Inverse Trigonometric, Functions, , 197-314, , 315-389, , Session 5, , •, , Property VI, VII and VIII of Inverse, Trigonometric Functions, , Session 6, • Property IX of Inverse, Trigonometric Functions, Session 7, • Property X, XI, XII and XIII of, Inverse Trigonometric Functions, PRACTICE PART, • JEE Type Examples, • Chapter Exercises, , www.jeebooks.in, Session 4, , •, , Property III, IV and V of Inverse, Trigonometric Functions

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www.jeebooks.in, i, 9, , ♦, Z, , Skills in Mathematics for, DKlllS Hl IVldLIlClIldllLb 1U1, , JEE MAIN & ADVANCED, , ■, , i, , I, , i, , I, , SYLLABUS FOR JEE MAIN, Trigonometry, Trigonometrical identities and equations. Trigonometrical, functions. Inverse trigonometrical functions and their, properties. Heights and Distances., , SYLLABUS FOR JEE Advanced, Trigonometry, Trigonometric functions, their periodicity and graphs, addition, and subtraction formulae, formulae involving multiple and sub­, multiple angles, general solution of trigonometric equations, , Relations between sides and angles of a triangle, sine rule,, cosine rule, half-angle formula and the area of a triangle, inverse, trigonometric functions (principal value only)., , www.jeebooks.in

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www.jeebooks.in, CHAPTER, , Trigonometric Functions, and Identities, Learning Part, Session 1, • Measurement of Angles, Session 2, • Definition of Trigonometric Functions, , Session 3, • Application of Basic Trigonometry on Eliminating Variables or Parameters and Geometry, Session 4, • Signs and Graph of Trigonometric Functions, Session 5, • Trigonometric Ratios of any Angle, Session 6, • Trigonometric Ratios of Compound Angles, Session 7, • Sum of Sines/Cosines in Terms of Products, Session 8, • Trigonometric Ratios of Multiples of an Angle, Session 9, • Trigonometric Ratios of sub multiple of an Angle, Session 10, • Trigonometric Ratios of the Sum of Three or More Angles, Session 11, • Maximum and Minimum Values of Trigonometrical Functions, , Practice Part, • JEE Type Examples, • Chapter Exercises, , www.jeebooks.in, : Arihant on Your Mobile!, , I Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.

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www.jeebooks.in, Session 1, , »«srm^r, , Measurement of Angles, The word ‘Trigonometry’ is derived from two Greek words., , (i) trigonon, , (ii) metron, , The word trigonon means a triangle and the word metron, mean a measure. Hence, trigonometry means measuring, the sides and angle of triangle. The subject was originally, develop to solve geometric problems involving triangle., , 2. Circular measurement or Radian measure The, angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle is, called a radian and denoted by lc., , Angle, In trigonometry, as in case of geometry. Angle is measure, of rotation from the direction of one ray about its initial, point. The original ray called the initial side and the final, position of the ray after rotation is called the terminal side, of the angle. The point of rotation is called the vertex. If, the direction of rotation is anti-clockwise, the angle is said, to be positive and if the direction of rotation is clockwise,, then the angle is negative, , 3. Centesimal or French system In this system of, measurement a right angle is divided into 100 equal, parts called Grades. Each grade is then divided into, 100 equal parts called minutes and each minute is, further divided into 100 equal parts called Seconds., , Thus, right angle = 100*, , 1° = 100', , l'=100", Note, , Initial side, 0 xJe, , o, , p, , Initial side, , (i) Positive angle, , (ii) Negative angle, , Angle of 90° is called a right anglel'of centesimal system * 1'of, sexagesimal system 1" of centesimal system * 1" of sexagesimal, system., This system of measurement of angles is not commonly used, and so here we will not study this system of measurement of, angles., , Measurement of Angles, , Radian is a Constant Angle, , There are three systems used for the measurement of, angles., , Let ABC be a circle whose centre is O and radius is r. Let, the length of arc AB of the circle by equal to r. Then by, the definition of radian., , 1. Sexagesimal system or English system (degree), 2. Circular measurement (radian), , 3. Centesimal system or French system (grade), , We shall describe the units of measurement of angle, which are most commonly used, i.e sexagesimal system, (degree measure) and circular measurement (radian, measure), 1. Sexagesimal or Degree measure If a rotation from, the initial side to the terminal side is (l/360)th of a, revolution, the angle is said to have a measure of one, degree, written as 1°. A degree is divided into 60, minutes, and a minute is divided into 60 seconds. One, sixtieth of a degree is called a minute, written as T;, one sixtieth of minute is called a second, written as 1"., Thus, 1° = 60' and T = 60, , Z.AOB -1 radian, Produce AO and let it cut the circle at C. Then AC is a, diameter of the circle and arc ABC is equal to half the, circumference of the circle., Also ZAOC = 2 right angle = 180°, , www.jeebooks.in, By geometry, we know that angles subtended at the centre, of a circle are proportional to the lengths of the arcs, which subtend them

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 3, , r, , Z.AOB arc AB, ----------- or, 180°, ZAOC arc ABC, , r, , 27tr, 2, , [•.• circumference of the circle = 2nr], , 180° 2 right angle, 1 radian =------= —-- ------- - — = constant, 71, 7t, , [since a right angle and 7t are constants], , Relation between Radians and Real, Numbers, .p, Consider a unit circle with center O. Let A, ■■2, be any point on the circle. Consider OA as, the initial side of an angle. Then the, -1, length of an arc of the circle gives the, 1 A0, radian measure of the angle which the arc, 0", subtends at the center of the circle., Consider line PAQ which is tangent to the, —2, circle at A Let point A represents the real, +Q, number zero, AP represents a positive real, number, and AQ represents a negative, real number. If we rope line AP in the counter-clockwise, direction along the circle, and AQ in the clockwise, direction, then every real number corresponds to a radian, measure and conversely. Thus, radian measures and real, numbers can be considered as one and the same., , Relation between Degree and Radian, It follows that the magnitude in radian of one complete, revolution (360 degree) is the length of the entire, 27tr, circumference divided by the radius, or----- or 2ti., r, , Therefore, 2n radian = 360°, or, , The Relation between Degree Measures and, Radian Measures of Some Common Angles, , Degree, Radians, , 30?, , 45°, , 6Cf, , 90?, , 180?, , it, , 71, , it, , 71, , 7t, , 6, , 4, , ,, c, ., 180, of the angle and degree measure ot an angle = — X, Tt, radian measure of the angle., , 2, , r, , I Example 1. Convert 40°21' into radian measure., Sol. We know that 180° = it radian., , 40°2T = 40- degree, , Hence, , 7t, 121, 12171, = — x — radian =------ radian., 180 3, 540, Therefore 40°21' =, , 540, , radian., , I Example 2. Express the following angle in degrees., .... (7ity, II - —, (i), , 112 J, (m) y, , ,. \, , 2n', , (,v) - —, , . . f( 571, _ . ..., 5ti Y, A (571, 57t 180, Sol. (i) — = — x —, it, 12, ” k112, 1277 112, , kO, , = (5X15)° = 75°, , 'lit 180Y, —x—, 12, it J, , or 1 radian =------ = 57° 16' (approximately), 7t, , 7t, Thus radian measure of an angle = —— x degree measure, , 2n, , e=-, , 180°, , 7t, :. 1° = — radian =0.01746 radian (approximately), 180, , 2, , 360?, , Note, (i) Radian is the unit to measure angle and it does not means, that n stands for 180°, it is a real number. Where as itc, stands for 180°., Remember the relation n radians = 180 degrees = 200, grade., (ii) The number of radians in an angle subtended by an arc of a, circle at the centre is equal to arc ., radius, , 7t radian = 180°, , Again, 180° = n radian, , 270?, , = -(7X15)° = -105°, , ) =19° 5' 27", , (ui) M =, , (1 180Y =, - X----:, <3, 7t ), , r \, , (2lt 180 lY, - —x — = -(2X20)° = -40°, I 9, 71, , (3J, , 2n‘, , (IV)- —:, , I Example 3. Express the following angle in degrees,, minutes and seconds form, (321.9)°, , www.jeebooks.in, Thus if the measure of an angle in degrees, and radians be, D and C respectively, then, , D, , C, , 180, , n, , Sol. (321.9)° =321° +0.9°, , = 321° + (0.9° X 60)', , = 321° + 54' = 321°54'

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www.jeebooks.in, II II, , 4, , Textbook of Trigonometry, , 3k = 3 X 32 = 96°, 4k = 4 X 32 = 128°, /. The other three angles measured in degree are 64°, 96°, and 128°., The angles in radians are, , I Example 4. In AABC, m ZA = —— and m Z B = 45°., Find m ZC in both the systems., Sol., , m ZA =, , 271, 180, — x — | =120°, 3, it, , 3, , m ZB = 45°, 7f, , = | 45 X —, , I, , 180, , I, , 180 J, , I, , 4, , It, , I, , x + y = 900°, x - y = 60°, On adding Eqs. (i) and (ii), we get, 2x = 960°, x = 480°, On putting x = 480° in Eq. (i), we get, 480° + y = 900°, y = 420°, Hence, the angles are 480° and 420°., , Sol., , I, , 180°, , I, , 71, , I, , 32?tc, 45, , ...(h), , ratio 2:3:4. Find their measures in radians and in, degrees., .o, , = 72°, , Since, measures of other three angles are in the ratio, 2 : 3 : 4. Let the angle be 2k, 3k and 4k measured in degree., /. Sum of all angles of quadrilateral = 360°, =>, 72° + 2k + 3k + 4k = 360°, =>, 9k = 288° => k = 32°, /.The other three angles are, 2k = 2 X 32 = 64°, , 3, 71, , 1071, , 180, , 3, , 600 X'-----, , It I, , (, , ...(i), , 2itc, , Tt, , (i) 120° = 120 x---I, 180, (ii) - 600° = -, , I Example 6. One angle of a quadrilateral has measure, 2nc, and the measures of other three angles are in the, , it ., , 8n', 15, , I Example 7. Express the following angles in radians., (i) 120°, (ii)-600°, (iii) -144°, , r c, C, 180 Y, x + y = 571 => x + y = 5n x---\, It J, , —x—, , it, , L°, , Sol. Let the angles be x and y in degrees., , .5, , 45, , \c, , .‘.The other three angles measured in radian are, 167tc 87tc, , 327tc, ----- ,---- and ------ ., 45 15, 45, , I Example 5. The sum of two angles is 5nc and their, difference is 60°. Find the angles in degrees., , 180, , 16n', , 128° = 128 x---I, 180., , mZC = —, 12, , 271, , =, , 96° = 96 x------, , mZC = 15 x---I, 180, , Sol. One angle =, , 7t Y, , z, , In AABC, m ZA + m Z.B + m Z.C = 180°, v The sum of angles of a triangle is 180°, 120° + 45° + mZC = 180°, 165°+ mZC = 180°, mZC = 180° - 165°, mZC = 15°, , Then,, , (, , 64° = 64 X-----, , (iii) - 144° = - 144 X---- =, I, 180 J, , 47te, 5, , I Example 8. If the three angles of a quadrilateral are, 5 it, , 60°, 60s and —. Then, find the fourth angle., 6, , Sol. First angle = 60°, , Second angle = 60* = 60 x, , 90, , degrees = 54°, , ,, ., 571, ,., 5X180, Third angle = — radian = —-— = 150°, Fourth angle = 360° - (60° + 54° + 150°) = 96°, , I Example 9. In a circle of diameter 40 cm, the length, of a chord is 20 cm. Find the length of minor arc, corresponding to the chord., Sol. Let arc AB = S. It is given that OA = 20 cm and chord, AB = 20 cm. Therefore, AOAB is an equilateral triangle., B, , Hence, ZAOB = 60°, , 20 cm', , = I 60 X —, I, , 180, , 60°, ', A, 20 cm, , www.jeebooks.in, Now,, , radius, 7t, S, „ 2O7t, — = — => S =---- cm, 3 20, 3

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , I Example 10. In the circle of 5 cm. radius, what is the, length of the arc which subtends and angle of 33°15z at, the centre., , Sol. Here, r = 5 cm; 15' = — =, 60, , n, 4], , 133, 0 = 33° 15' = 33 + — = — degrees, 4, 4, 133, 71, 133, 22, 1463, ,., ~ — x — = — x-------- =------ radians, 4, 180, 4, 7 X 180 2520, Now, 0 =r, ,, 1463 _ 65, I = 0r =------x 5 =2— cm (approx.), 2520, 72, , I Example 11. The minute hand of a watch is 35 cm, long. How far does its tip move in 18 minutes?, T, 22^, , 1), , I, , Sol. The minute hand of a watch completes one revolution in, 60 minutes. Therefore the angle traced by a minute hand, in 60 minutes = 360° = 2tc radians., .'. Angle traced by the minute hand in 18 minutes, 18, , ,., , 3tc, , 5, , = 35 x — = 2171 = 21 x — = 66 cm, 5, 7, , I Example 12. The wheel of a railway carriage is 40, cm. in diameter and makes 6 revolutions in a second;, how fast is the train going?, So/. Diameter of the wheel = 40 cm, , :. radius of the wheel = 20 cm, Circumference of the wheel = 2itr = 2n x 20 ~ 40it cm, Number of revolutions made in 1 second = 6, A Distance covered in 1 second = 4071 x 6 = 2407t cm, A Speed of the train = 24071 cm/sec., , I Example 13. Assuming that a person of normal sight, can read print at such a distance that the letters, subtend an angle of 5' at his eye, find the height of, the letters that he can read at a distance of 12 metres., Sol. Let the height of the letters be h metres., Now, h many be considered as the arc of a circle of radius, 12 m, which, subtends an angle of 5' at its centre., , ,., , = 2it x — radians = — radians, 60, 5, Let the distance moved by the tip in 18 minutes be I, then, , '___7C, , 0 = 5'= [ — X, | radians =, radian, 160 180 J, J2X180,, and, , r = 12m, 71, , ( it, , h = r0 = 12x—-— = — metres = 1.7 cm, 12 X 180 1180 J, , I=r0, , Exercise for Session 1, 3n, , 1. The difference between two acute angles of a right angle triangle is — rad. Find the angles in degree., 10, 2. Find the length of an arc of a circle of radius 6 cm subtending an angle of 15° at the centre., , 3. A horse is tied to post by a rope. If the horse moves along circular path always keeping the tight and describes, 88 m, when it has traced out 72° at centre, find the length of rope., 4. Find the angle between the minute hand and hour hand of a clock, when the time is 7:30 pm., 5. If OQ makes 4 revolutions in 1s, find the angular velocity in radians per second., 6. If a train is moving on the circular path of 1500 m radius at the rate of 66 km/h, find the angle in radian, if it has, in 10 second., , 7. Find the distance from the eye at which a coin of 2.2 cm diameter should be held so as to conceal the full moon, with angular diameter 30'., 8. The wheel of a railway carriage is 40 cm in diameter and makes 7 revolutions in a second, find the speed of train., 9. Assuming that a person of normal sight can read print at such a distance that the letters subtend an angle of 5', at his eye, find the height of letters that he can read a distance of 12 m., , www.jeebooks.in, 10. For each natural number k, let CK denotes the circle with radius k cm and centre at origin. On the circle CK, a, particle moves k cm in the counter-clockwise direction. After completing its motion on CK, the particle moves on, CK +, in the radial direction. The motion of the particle continues in this manner. The particle starts at (1, 0). If, the particle crosses the positive direction of the x-axis for the first time on the circle C„, then n is equal to

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www.jeebooks.in, Session 2, , Definition of Trigonometric Functions, , Definition of Trigonometric, Functions, An angle whose measure is greater than 0° but less than, 90° is called an acute angle., In a right angled triangle ABC, Z.CAB = A and Z.BCA = 90', = 7i/2. AC is the base, BC the altitude and AB is the, hypotenuse. We refer to the base as the adjacent side and, to the altitude as the opposite side. There are six, trigonometric ratios, also called trigonometric functions, or circular functions with reference to ZA the six ratio, are, 8, , Hence, |sin A | < 1, cos A | < 11, |cosec A | > 1, [sec A | > 1,, while tan A and cot A may have any numerical value lying, between - °° to +, , Note, Student must remember the following, (ii), (I) -1£sin4«Sl, (iv), (iii) cosec A > 1 orcosec A < -1, (vi), (v) tan/le/?, , Some values of Trigonometrical Ratios, Students are already familiar with the values of sin, cos,, tan, cot, sec and cosec of angles 0°, 30°, 45°, 60° and 90°, which have been given in the following table, , BC, , C, , opposite side, is called sine of A, and written as, hypotenuse, , AB, sin A, AC adjacent side, AB, hypotenuse, as cos A, BC opposite side, , is called the cosine of A, and written, , is called the tangent of A, and written, AC adjacent side, as tan A, AB, hypotenuse, is called cosecant of A, and written as, BC opposite side, cosec A, AB, hypotenuse, is called secant of A, and written as, AC opposite side, sec A, AC adjacent side, is called cotangent of A, and written, BC opposite side’, as cot A, Since, the hypotenuse is the greatest side in a right angle, triangle, sin A and cos A can never be greater than unity, and cosec A and sec A can never be less than unity., , 30°, , o, , r, , cos, , 1, , tan, , 0, , /3, 2, 1, /3, , cot, , undefined, , sin, , A, , results, -1 < cos A 51, sec A >1 or sec -A £ -1, coMefl, , 2, , 1, , sec, , cosec undefined, , 45°, , jZ, V2, i, V2, , /3, , 6QP, , 71, , T, 2, 2, , 3, _1_, , 1, , 9(F, 1, 0, , undefined, 0, , V3, , ~T_, 43, , 2, , 2, , 2, , 2, , 2, , undefined, , 1, , /3, , Trigonometric Identities, Trigonometric identities are equalities that involve, trigonometric functions that are true for every single, value of the occurring variables. In other words, they are, equations that hold true regardless of the value of the, angles being chosen., , Trigonometric identities are as follows, 1. sin2 A+cos2 A = 1 => cos2 A = l-sin2 A, , or, , sin2 A = l-cos2 A, , 2.1 + tan2 A = sec2 A =» sec2 A-tan2 A = 1, , www.jeebooks.in, 3. cot2 A +1 = cosec2 A, , cosec2 A - cot2 A = 1

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www.jeebooks.in, Chap 01, , Trigonometric Functions and Identities, , 11, , Sol. From Fig. we have, , Sol., , it, «?_ tan[ ---0, PR, 12 ), , Y, 8, , RS, = cot 0 and — = tan 0, PR, S, , A, , X, 10, , x, , 1, , Q, , We have area of AXYB = - area of AABC, 2, .. 1, -(XY)-(XB) = -x-x AC x BC, 2, ' 2 2, (x x, ^ = 24, \ 2 J” 2, or, or, , or, , x x x tan B = 24, - 3, x2 X - = 24, 4, x2=32 orx = 472, , X, , f-e, , 9, , -I--------------, , P, , [v y = x tan B], , R, , 0, , [•.•tanB = —], BC, , I Example 31. Let PQ and RS be tangents at the, extremities of the diameter PR or a circle of radius r. If, PS and RQ intersect at a point X on the circumference, of the circle, then prove that 2r=y]PQ xRS., , or, , PQ RS, —x—=1, PR PR, (PR)2 =PQx PS, , or, , (2r)2 = PQ x PS, , or, , 2r = fPQ x PS, , Exercise for Session 3, 1. If sec 0+ tan Q = k, find the value of cos a, 2. If x sin3 0 + y cos3 0 = sin 0cos 0and x sin 0 = y cos 0,, , Find the value of x2 + y2., 3. If sin A +cos A = m and sin3 A + cos3 A=n, prove thatm3 -3m + 2n =0., , 4. If sin2 0 =, , x2 + y2 + 1, . Find the value of x and y., 2x, , 5. If sin 0 -76 cos 0=77 cos 0. Prove that cos 0 + 76 sin 0 - 77 sin 0 = 0., 6. If sin x + sin y + sin z = 3. Find the value of cos x + cos y + cos z., 7. If — cos 0 + — sin 0 = 1, — sin 0 - — cos 0 = 1 then eliminate 0., a, b, a, b, , 8. If a sin2 x + b cos2 x = c, b sin2 y + a cos2 y = d and a tan x =b tan y, then prove that, , ((j *** 3 ) (c_ 3 ), , = 7------ —-f., b* (b-c)(b-d), , 9. If a + b tan 0 = sec 0 and b -a tan 0 = 3 sec a then find the value of a2 + b2., , www.jeebooks.in, 10. Two circles of radii 4 cm and 1 cm touch each other externally and 8 is the angle contained by their direct, , 6, , 0, , common tangents. Find the value of sin - + cos -.

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www.jeebooks.in, Session 4, Signs and Graph of Trigonometric Functions, , Signs of Trigonometric, Functions, The signs of the trigonometric ratios of an angle depend, on the quadrant in which the terminal side of the angle, lies. We always take OP = r to be positive (see figure)., Thus the signs of all the trigonometric ratios depend on, the signs of x and/or y., Y, , X'-, , X', , ■X, , Y, <, An angle is said to be in that quadrant in which its, terminal ray lies, , For positive acute angles this definition gives the same, result as in case of a right angled triangle since x and y, are both positive for any point in the first quadrant and, consequently they are the length of base and, perpendicular of the angle 0., Y, Second quadrant First quadrant, (sin, cosec, (all are positive), are positive), , X', , ---------------- "X, O, Third quadrant Fourth quadrant, (cos, sec, (tan, cot, are positive), are positive), , r, 1. Clearly in first quadrant sin 0, cos 0, tan 0, cot 0, sec 0, and cosec 0 are all positive as x, y are positive., , 4. In fourth quadrant, x is positive and y is negative,, therefore, only cos0 and sec 0 are positive., Quadrant —>, sin 0, , I, +, , COS0, , +, , tan0, , +, , cosec 0, , +, , sec0, , +, , cot0, , +, , II, +, , IV, , III, , +, , +, +, +, , +, , Variation in the Values of, Trigonometric Functions in Different, Quadrants, We observe that in the first quadrant, as x increases from, 0 to —, sin x increases from 0 to 1 and in the second, 2, 7C, quadrant as x increases from — to 7t, sin x decreases from, , 1 to 0., , 371, In the third quadrant, as x increases from 7t to —, sin x, decreases from 0 to -1 and finally, in the fourth quadrant,, 3 7t, sin x increases from -1 to 0 as x increase from — to 271., 2, , 2nd, . 3rd quadrant 4th quadrant, quadrant, ? from 0 to 1 X from 1 to 0 J- from 0 to ? from - 1 to, -1________ 0_________, i from 1 to 0 -1 from 0 to ? from -1 to ? from 0 to 1, -1, 0, _______________, T from 0 to oo T from - °°, ? from 0 to 00 ? from - 00, , Function 1st quadrant, sin 0, , cos 0, , tan 0, , to 0, , cot 0, , 2. In second quadrant, x is negative and y is positive,, therefore, only sin 0 and cosec 0 are positive., , sec 0, , 3. In third quadrant, x and y are both negative,, therefore, only tan0 and cot 0 are positive., , cosec 0, , from 00 to 0, , from 0 to, , to 0, , •I from 00 to 0 i from 0 to, , — 00, , — 00, , ? from 1 to OO T from - 00 X from - 1 to X from °° to 1, to -1, I from 00 to 1 T from 1 to 00 ? from - °° X from - 1 to, , www.jeebooks.in, to -1, , — oo

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 13, , Note, , + «and - ~are two symbols. These are not real numbers. When, n, we say that tan 0 increases from 0 to « as 0 varies from 0 to it, 2, means that tanO increases in the interval (o,, and it attains, , 3kT, 2:, , arbitrarily large positive values as 0 tends to —. This rule applies to, other trigonometric functions also., , Graphs of Trigonometric, Functions, , ( FT s, , I 2, , 4. y=f(x) =cot x, Domain —> R - rm, nE I; Range —> (-, , n,, , °°); Period, , x, , As in case of algebraic function, we can have some idea, about the nature of a trigonometric function by its graph., Graph has many important applications in mathematical, problems. We shall discuss the graphs of trigonometrical, functions. We know that sin x, cos x, sec x and cosec x are, periodic functions with period 2n and tan x and cot x are, trigonometric functions of period K. Also if the period of, T, function /(x) is T, then period of /(ax +b) is —., , i2it, , +-X, , kl, , 71, , Graph and Other Useful Data of, Trigonometric Functions, 1. y =/(x)=sinx, , Domain —> R ~ (2n +1) —, n. E I, 2, Range —>(-<»,-l]u[l,<»), Period 2n, sec2 x, | sec x | G [1, ~), , r, , Domain —> R,, Range—>[-1,1], Period -> 2n, y, , -it, , 0 ii/2, , it, , ~3n/2, , Ts, , o, , 71, , ..2*, , i 3* 2n, '2____, , •2.., , /--------- H—►X, , 271 571/2, i, , 2. y=f(x) =cosx, Domain —> R, Range —> [-1,1], , Period —> 2n, , 6. y =/(x) = cosec x, Domain —> R ~ rm, n e /;, Range —>(-°°, -l]u[i,o°), Period —> 2k, cosec2 x, | cosec x | e [1, °°), y, I, I, , 1, I, I, , 3. y =/(x) = tan x, , it, , 0, , 7t, , ]2T, , *x, , www.jeebooks.in, 71, , Domain -> R ~ (2n +1) —, n e I, 2, Range (- 00), Period->k, , 2., , 2, , I, , I, I

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www.jeebooks.in, 14, , Textbook of Trigonometry, , Transformation of the Graphs of, Trigonometric Functions, , y, , y=sin (x/2), I, , 1. To draw the graph of y = f(x +a); (a > 0) from the, graph of y = /(x), shift the graph of y = /(x), a units, left along the x-axis., Consider the following illustration., y, , x, , y=sinx, , n/2l, , 7t?-..37C/2;, , y =|sin x|, 1, , ^2tt*X, , I, , ., I /, \ 1 /, , To draw the graph of y = f(x - a); (a > 0) from the, graph of y = /(x), shift the graph of y = f(x), a units, right along the x-axis., Consider the following illustration., 4, -27?/<37T/2k -n|, , y - cos x y = cos, , 1, , i, , i, , ft, , I-tt/2 / 71/2X2, , 1, , 1, , I, , n/2;, , -ti/2;, , i, , i4n, , 1, 2, , I, , i, , fT-7---- ♦’X, , = 4rt, , Period of y =sin( — is, , I, I, , 77X377/2!, , <7jt/2, , 1577/2 b.37C X, , X, 7^, , ------------ 1------------ r-, , \!, , !, , y=sin (x+1), /, ,y=sinx, I, , I, , 1, , I, I, , 2rc;, , 377/2;, , 77;, , *x, , 4. Since y =| f(x) | > 0, to draw the graph of y = | f(x) |,, take the mirror of the graph of y = /(x) in the x-axis, for /(x) < 0, retaining the graph for f(x) > 0., Consider the following illustrations., Here, period of /(x) = | sin x | is it., yK, , ", 3n/2 1/271, j, , y =|cosx|, , ■x, , ., , 71, , ., , 1, , V--- T", , 3, , I, , I, , 2. To draw the graph of y = /(x) + a\ {a > 0) from the, graph of y - f(x), shift the graph of y = /(x), a units, upwards along the y-axis., To draw the graph of y = /(x) - a; (a > 0) from the, graph of y = jf(x), shift the graph of y - f(x), a units, downward along the y-axis., y, 1, 1, 1, 1, 2J, T--4, •, i y=cosx+1, , -7c], , -77/2 !, , 77/2], , 371/2;, , *x, , Here, period of /(x) = | cos x | is it., y=|tan x|, 7 “, f, , l, , j, , ’/a72=cosx, , 7C[-37t/^, , -77;, , /j-jt/2, , 1377/2, i2rtX, r, , rt/2;, , -J______, , 1, , 4-----, , - - -4, , ; y=cosx-2, , . _ j________ -, , 4, , I ~2, , ■, , I, , r, , “ T “ “ ’, , 3. If y =f(x) has period T, then period of y = f(ax) is, T, , I, , l4, -271 A, , -, , 3}, , jl, , /, , r*\/1, , 1 kCy 71, ZT'C, 1 v, 1, VI, , 2, , ■>x, , 1, , •^7t-i/2) 3^2, Period of y = sin(2x) is — = it, 2, , /, , ', , -2, , 5. Graph of y = af(x) from the graph of y f(x), yt, y=2sinx, , y= sin (2x), y= sin x, , ..J 4, , ;, , 1 I, , i, , y=tan x —■—►/, , ■►X, , “, 3tc/2<, , /iTC, Thi, , i77/2, aU2, , i-7t/2, , "i ”, , >, 1, , 1, , -* 1r -* -" -"1 -” 1 1, 1, , 1 !/, , I, , I, , 1, , 77/2 /, , ± y=3(sinx, , <, , 1, , X, , i, , __ i, , \\ 1, , '.K 57^/2377^, II, I, , 1/, I, , Z.2-----, , «yr=sin x, , >x, , I, , I, , www.jeebooks.in, °, , I, , I

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , y=2secx, / y, , I, I, , I, I, I, , T, , r, , I, i, I, i, , hi/2, , ., , I, i, , y, , ■, , hi, , I Example 32. Find the values of the other five, trigonometric functions in each of the following, questions, , y = secx, , i, I, i, , i, i, i, , i, l, i, , \, , i, , t, I 3rt/2 J, , 7U2, , —zc/2, , (i) tan 0 =, , where 0 is in third quadrant., , (ii) sin 0 =, , where 0 is in second quadrant., , i, , iiii, , !, , 15, , 71, , wr;. Wi, , -i—*x, 5n/2, I, I, , 2, , " t j " " t----yi" “ 4------ r;-p - -i- - y. r - —r, , So/. (i) Since 0 is in third quadrant,, Only tan 0 and cot 0 are positive, tan 0 = —, 12, , Now,, I, , r*, , cot0 = —,, 5, 5, sin 0 = 13, 13, cosec = ----5, 12, 13, cos0 =---- and sec 0 = 12', 13, (ii) Since 0 is in the second quadrant,, Only sin 0 and cosec 0 will be positive., , Therefore,, , Some Important Graphical Deductions, To find relation between sin x, x and tan x, y=x, , (i), , 2, , sin x'—, , sin 0 = -., 5, , Now,, Thus, when, , Therefore,, , <x<0, , => sin x > x, (ii), , y, , ■]tanx=y, , ;, , >y=x, , and, , 1 ■, , hi, , 4, , • i, , “O, , >X, , A 'A, 4 |2, , Sol. We have sin2 0 + cos2 0 = 1, , tan x > x, when 0 > x > —, 2, , y, , 12, , I Example 33. If sin 0 = — and 0 lies in the second, , quadrant, find the value of sec 0 + tan 0., , I, I, , (iii) In general,, , A 5, A, 4, cosec 0 = -, cos 0 - —,, 3 5, 5, 3, sec 0 = —, tan 0 = —, 4, 4, 4, cot 0 = —., 3, , cos 0 = ±y/l - sin2 0, , =>, , In the second quadrant, cos 0 is negative, cos 0 = - -Ji - sin2 0, , tanx, , y=x, 1, sin 0 1 + sin 0, Now, sec 0 + tan 0 = ----------- -]------------- = ., cos 0 cos 0, cos 0, , hi, , 4, , sinx, , 0, , 1+—, 13, , 1 + sin0, , >X, , ra, , - -Jl - sin2 0, , 113 J, , ( 7C, Thus, tan x > x > sin x, V x G 0, —, , www.jeebooks.in, I 2, , JI 1, and sin x > x > tan x, V x G -—,0 L, , 25, , „, , 13, , I, , I 2 J, , V169, , 25, , _ -13- =-5, , £, , 13

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www.jeebooks.in, 16, , Textbook of Trigonometry, , I Example 34. Draw the graph of y = 3sin 2x., , Sol. sin x is a periodic function with period 27t, therefore,, 2n, sin 2x will be a periodic function of period — = 7t, , — 1 < sin 2x < 1, -3<3sin 2x <3, In order to draw the graph of y = 3sin 2x, draw the graph of, k, k, y = sin x and on X-axis change k to —, i.e. write — wherever, 2, 2, it is k. For example, write 15° in place of 30°, 45° in place of, 90° etc., On Y-axis change k to 3k, i.e. write 3k wherever it is k for, example, write 3 in place of 1, - 3 in place of - 1, 1.5 in place, of 0.5 etc., The graph of y = 3 sin2x will be as given in the figure., , In order to draw the graph of y = cos x - — , we draw the, 4J, graph of y = cos x and shift it on the right side through a, 71, , distance of — unit., 4, , Also, , X, y=3 sin 2x, , 3, -it, , n n\ 3n, 4 2 \4z, , n\ n; 0/, , it, , 3n, 2, , I Example 36. Which of the following is the least?, (a) sin 3, (b) sin 2, (c) sinl, (d) sin 7, Sol. (a) sin 3 = sin[n - (it - 3)] = sin(7t - 3) =sin (0.14), sin 2 = sin[7t -(it - 2)], = sin(7t - 2) =sin (1.14), sin 7 = sin[27t +(7 - 2n)], = sin(7 - 2tc) =sin (0.72), Now,, 1.14 > 1 > 0.72 > 0.14, =>, sin (1.14) > sin 1 >sin(0.72) >sin(0.14), [as 1.14, 0.72, 0.14 lie in the first quadrant and sine, functions increase in the first quadrant], , Hence, among the given values, sin 3 is the least., Alternate solution, X, , I Example 35. Draw the graph of y =cos, Sol. Given function is y =cos, , 7t, , X------, , n, , : 2, , w, -1, , 4), , !n, , I, , j, , 37^, , I, , 2!, , I Example 37. Find the value of x for which, J(x)=^sin x-cos x is defined, xe [0,2tc]., Sol. f(x) =y/sin x - cos x is defined if sin x > cos x., , Jt, , x=—, 4, , y=sinx, , y, ■, , ", , I, , I, , I /, , y=cos(x- J), , I, I, , ■, , I, I, , I, , 5n, 4, , 9n, 4, , I, , i, , I, I, I, , \L, , 1, 1, i------- r------ 1, ■, i, i, , 1 X 1, , 7^4 7t/$, , -7^/4, , y=cos x, , • J ■, , ■, , I, , 3n, >4, , 7 ', 1 5tc/2, , L, , From the graph, obviously sin 3 is the least., , Xf, , O IE, 4, , I, I, , 3x/2, , __1__, , 7t, , it, , 1, , I, I, I, , I, , 71 ], X-----, , X = x---- and Y = y, 4, Y=0 => y = 0 and X = 0, x---- = 0 =>, 4, , 1, , 4, , Given function is Y = cos X, where, , or, , ■, , a, , 1, , X1, , \i1, , r, , I, , I, , 1x, , 1, , ’z, , 1, , 1, , 1, , 3n/4 n\57t/4/^t/2 7n/j, 1, , 1, I, I, , t, , 1/, , I, I, r ~’, , I, , X, , 7t 5n, From the graph, sin x > cos x, for x e —, •—, 4* 4, , www.jeebooks.in

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www.jeebooks.in, Chap 01, , Trigonometric Functions and Identities, , 17, , I Example 38. Solve tan x > cot x, where x e [0, 2jc]., , 4, , Sol., , T-4 'y= cotx-, , •r-3-, , . 2 ..., , -■l-y= tanx'r, , - 1 --, , n/4, 1, , jt/4, , rcjX 3n>4, 1 \ • ., , •x, , n, I, , 5n/4 5nJ, I, , t, , 7nJ4/2iz, ., I, /, I, , - 4 --, , -3-, , i, , r-2--, , I---I----- 1, , We find that tan x > cot x. Therefore, the values of tan x, are more than the value of cot x., That is, the value of x for which graph of y = tan x is above, the graph of y = cot x., , From the graph, it is clear that, XG, , it, 4’ 2, , (lit, , >, , f 57t, , f?7t, , 'l, , U ) U 2) U, , J, , Exercise for Session 4, 4 3, 1. If tan x = - -, ~ < x < 2tc, find the value of 9 sec2 x - 4 cot x., <1, 2. Show that sin2 x = p + - is impossible if x is real., P, , o .<•, __ _x = 3 and x lies in the fourth quadrant find the values of cosec x + cot x., 3., If cos, 5, X, , 4. Draw the graph of y = sin x and y = sin —., " ', '2, , 5. Draw the graph of y = sec2 x - tan2 x. Is f(x) periodic? If yes, what is its fundamental period?, , 6. Prove that sin 0 < 6 < tan 0 for 0 e 0, — ., , I 2J, , 7. Find the value of x for which f(x)=A/sin x - cos x is defined, x e [0,2n]., 8. Draw the graph of y = sin x and y =cos x, 0 < x < 2n., , 9. Draw the graph of y = tan(3x)., , www.jeebooks.in, ft, , 10. If cos x - ---------and — < x < n, find the value of sin x., 4, 2

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www.jeebooks.in, Trigonometric Functions and Identities, , Chap 01, , Allied angles (or numbers), , Two angles (or numbers) are called allied iff their sum or, 7T, , 7T, , TC, , difference is a multiple of —. For example, — and — are, 2, 3, 6, , 19, , 71 ± 0 j + i sin | — ± 0, => i • (cos 0 ± i sin 0) ~ cos \ —, \2, 2 J, , (it, A, 7C, => i-cos0+sin0 =cos — ±0 + isin| — ±0, 2, U J, 7, , On comparing real and imaginary part of LHS and, RHS, we get, , allied, — and - — are allied., 6, 6, , AID TO MEMORY, , ^71, , cos — -0^=sin0, , cos —+0 =-sin0, , U J, , You must have been overwhelmed by large number of, formulae for allied angles (or numbers). Instead of memorising, all of them, use the following rules, 1. Any trigonometric function of a real number nn ± x(n e /),, , U, , sin! — +0 I =cos0, , sin---- 0 = cos 0, , U, , J, , 12, , ), , J, , 7C, , treating x as 0 < x < -, is numerically equal to the same, function of x, with sign depending upon the quadrant in, which the arc length (on the unit circle) terminates. The, proper sign can be ascertained by ‘All - Sin - Tan - Cos’, rule. For example, sin(n + x) = - sin x; - ve sign was, chosen because n + x lies in the third quadrant and sin is, - ve in the third quadrant., It, , 2. Any trigonometric function of a real number (2n +1) - ± xm, , II. Method, To prove cos(rc ±0) =-cos0 and sin(7t ±0) = +sin0, , el(n±0) = cos(7t ±0)+ isin(7t ±0), , Since,, , •e‘(± 0) =cos(n ±0) + isin(7t ±0), =^> -(cos(±0) + isin(±0)) =cos(n ±0) + isin(n ±0), , On comparing real and imaginary part, we get, , treating x as 0 < x < ^, is numerically equal to cofunction, , cos (it +0) = -cos0, , of x, with sign depending upon the quadrant in which the, arc length (on the unit circle) terminates. Note that sin and, cos are cofunctions of each other; tan and cot are, cofunctions of each other; sec and cosec are cofunctions, , of each other. For example, seel - + x I = - cosec x, - ve, , cos(n — 0) = —cos0, , sin (ft +0) = -sin0, sin(7t -0) =sin0, , I Example 39. Prove that, , 7t, , sign was chosen because - + x lies in the second, , quadrant and sec is - ve in the second quadrant., , (i) sin2 —+cos2 —-tan2, 63, .... _ . i17t, 7t, , 71, , 1, , 4, , 2, , 3, , , 7T, , i 77t, 7 7t, , (ii) 2sm —+cosec —cos — = ’2, (n), 6 6, , I. Method, ( 71, , To prove cos — ±0 = +sin0, , U, , J, , and sin — ±0 =cos0, , 1, , ", , <2, , 36, , (m) cot —+cosec—+3 tan — = 6, 6, , 6, , 6, , (iv) 2sin2 —+2cos2 —+2sec2 — = 10, , Proof, , 2, , 2, , Sol. (i) We have,, , J, , \, £, = cosf— ±0 + isinf—, 7t ±0, => e 2 •e±,e, , u J, , 2, , 7t, 'I, i 71, /v, =>• i-e‘(±e) = cos —±0 I+ isin| —, ±0, 2, , 3, , 4, , 4, , if-±e] = cosf—, 7t ±0^ +isinf—±0, , 7, , • 2 7t, , . 7t, , 1, , 2, , sin —+ cos----tan — =, 6., 3, 4, 2, , n, • —, n + I cos —, = I sin, I, 3, 6., 1, , 12,, , 2, , +, , 1 1, -+4 4, , 1, , 2, 2, , n, tan —, 4, , 2, , I ’(I)2, , 2, , 1, , 1, , 2, , 1, , 1, 2, , www.jeebooks.in

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www.jeebooks.in, 20, , Textbook of Trigonometry, , (ii) We have,, „ . 27t ,, , 27n, , 2sm — +cosec----- cos, 6, 6, /, , . It, sin —, , X2, , I, , 6, , 771, , I, , _(-sin 0)(sec0)(- tan 0), (sec 0) (- sin 0) (tan 0), , 2, , n, , cos —, 3, , 6, , = -l, = R.H.S., , 2, , 71, , I, , 7C, COS —, , cosec Hr + —, k, 6, , 6, , 71, , 2, , + cosec —, , „ I . 71, = 2 sin —, , J •, , 3, , /, , 6, , I, , Sol L H s cos(90° + 9) sec(~ 9) tan(180° - 0), ' sec(360° - 0)sin(18O° + 0) cot(90° - 0), , 27t, , n, -cosec—, 6, , = 2 sin —, , 3., , I Example 41. Show that tan1° tan2°...tan89° =1, , \2, , ■> 2 /, , 7t 1, COS —, , Sol. L.H.S. = (tan 1° tan 89°) (tan 2° -tan 88°)..., , 3J, , = [tan 1° tan(90° - 1°)] - [tan 2° tan(90° - 2°)], ... [tan 44° tan(90° - 44°)] tan 45°, = (tan 1° • cot 1°) (tan 2° • cot 2°), ... (tan 44° ■ cot 44°) tan 45°, [•/ tan 0 cot 6 = 1 and tan 45° = 1], =1, , [•.• cosec(7t + 0) = -cosec0], 2, , 1, 2, , =2, , T=i, +I=i2, >), 2, , I +(-2)2x, , (iii) We have,, .2 7t, , 571, , 2 71, , cot —F cosec — + 3 tan —, 6, , 6, /, , = I cot—, k, 6, , I Example 42. Show that, , 6, , \2, , I, , 71), , I, , 6, , „f, , + cosec Hr---- +3 tan —, , = (V3)2 4-24-3, , 7t, 71, , 2, , sin2 5° + sin210° + sin215° + ...+sin2 90° = 92, , 6, , 2, , 1, , +, , So/. L.H.S. = (sin2 5° + sin2 85°) + (sin210° +sin2 80°) +, (sin2 40° + sin2 50°) + sin2 45° + sin2 90°, , A, , = (sin2 5° +cos2 5°) +(sin210 + cos210°), , =3+2+1=6', , + ... + (sin2 40° + cos2 40°) + sin2 45° +sin2 90°, , (iv) We have,, „ - 2 371, , 2 71, , _, , 2 7t, , 1, , 2sm — + 2 cos — + 2sec —, 4, 4, 3, n( . 371 ), , 2, , z, , J, , 71 1, , /, , 7C, , I, , 4, , \2, , /, , \2, , 2/X 2, , = 2 sm —, , 71, , + 2 cos —, , 4, , /, , 7t, , = 91, 2, , \2, , -2 sec—, , 3tt, .371, , I Example 43. Find the value of, , 3, ., , 71 1, , .71, , '.'sm— = sin | 71---- | = sm —, 4, 4, 4, =2, , 1, , Ji I, , 2, , +2, , 1, , Ji, , 2, , 2 71, 2 311, 2 6 71, 2 F 71, — + COS ---- + COS ----- + COS —, , 16, , 16, , 16, , 2 71, , 2 371, , 16, , 71, , 3ti, , 2, , 16, , + cos, , 2, , n n_, 2, , 16, , , 2 71, , , 2 371, , = cos — + cos— + sin — +sin —, 16, 16, 16, 16, 16, , = 1 + 1 + 8 = 10, , sec(360° - 0) sin(180° + 0)cot(90° - 0), , COS, , Sol. L.H.S. = cos2 — + cos2 — + cos2, 16, 16, , I +2(2)2, , I Example 40. Prove that, cos(90° + 0)sec(- 0) tan(180° -9), , +1, , Ji, , 71), . = 2 sm — + 2 cos — I + J, 2 sec—, k, 4j1, k, 4 J, I 3j, J ■, , 2, , I, , 2 7t ., , 2 71, , = cos — + sm —, I, 16 ., 16, , = -1., , + cos2 — + sin, 16, , 2 371, , 16, , =1+1=2, , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 21, , Exercise for Session 5, 1. Find the value of tan----- ., 3, , -, , 2. Find the sign of sec 2000°., 3. The value of cos f + cos 2° + cos 3° +... + cos 180°., , 4. Find the value of cos(270° + 0)cos(9O° - 0)-sin(27O° - 0)cos 0., 5. If Sn = cos" 0 + sin" 0, find the value of 3S4 - 2S6., , 6. sin2 0 =, , x2 + y2 + 1, 2x, , , then x must be., 1, , 7. If sin x + cosec x = 2, then find the value of sin'i10 x + cosec10 x., x, 8. e*sintax, -e“sinx=4 then find the number of real solutions., , 9. If n < a <, , then find the value of expression ^4 sin* a + sin2 2a + 4 cos2, 4, , n, , 2, , n, , 10. If J cos 0, = n, then the value of £sin 0,., /=i, , Session 6, Trigonometric Ratios of Compound Angles, Trigonometric Ratios of, Compound Angles, Algebraic sum of two or more angles is called a compound, angle. If A, B, C are any angles then A + B, A - B,, A + B+C, A-B+C.A-B-C, A + B-C, etc., are all, compound angles., Till now, we have learnt the values of trigonometric ratios, between 0° to 360°. Now, we are going to learn the values, of trigonometric ratios of compound angles., Note, Trigonometric ratios if i.e. sine, cosine, tan, cot, sec and cosec, are not distributed over addition and substraction of 2 angles., i.e., sin(4+8) *sin4+sinB, Proof:, A = 60°, B = 30°, sin(90°) # sin60°+sin30°, , (ii) cos(A + B) = cos A cos B - sin Asin B, tan A + tan B, (iii) tan(A + B) =, 1 - tan A tan B, vZ, , OK, , Qlf, /a, , A, , y, , H, , -90r 'p, A, , e/, , O, , M L, , x, , M, , W, O, , L, , ■x, , Let the revolving line starting from the position OX, describe first Z.XOY = A and then proceed further so as to, describe Z.YOZ = B in its position OZ., Then,, ZXOZ = A + B, In figure 6.1 A + B < 90° and in figure 6.2 A + B > 90°, Let Q be a point on OZ. From Q draw QM1 OX and, QP1OY. From P draw PH 1 QM., Now,, Z.HPO = Z.POX = A, , www.jeebooks.in, The Addition Formula, , (i) sin(A+B) =sin A cos B +cos AsinB

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www.jeebooks.in, Chap 01, , Trigonometric Functions and Identities, , 23, , Comparing real and imaginary parts of the left and right, hand side, we get,, , 3. sin(A+B) = cos —-(A + B), 2, , cos( A ± B) = (cos A cos B + sin A sin B), , = cos, , —-A -B, , = cos — - A | cos B + sin |, , 1, I, , 2, , I, I, , 4. tan(A + B), a, , sin(A ± B) = (sin Acos B ±cos Asin B), , I2 ) ., r, , n - A | sin B, 2, , = sin A cos B +cos A sin B, sin(A + B), , TWO VERY IMPORTANT IDENTITIES, , (iii), , (a) sin (A+8)-sin(A-8), = sin2 A -sin2 B = cos2 B -cos2 A, , (b) cos (A+8)-cos(A-8)=cos2 A-sin28, , cos(A +B), , sin Acos B +cos A sin B, cos Acos B - sin Asin B, , tan A + tan B, , ,(iv), , 1 - tan A tan B, , Proof: (a) sin(A + B)-sin(A-B), = (sin Acos B + cos Asin S)(sin Acos B -cos Asin 8), = sin2 Acos2 B -cos2 Asin2 8, = sin2 A(1 -sin2 8) -sin2 8(1 -sin2 A), = sin2 A-sin2 8, , (b) cos(A+8)-cos(A-8), = (cos A cos 8 -sin Asin 8) (cos Acos 8 + sin Asin 8), = cos2 Acos2 8 -sin2 Asin2 8, = cos2 A(1 -sin2 8) -(1 -cos2 A)sin2 8, , [dividing numerator and, denominator by cos A cos B], 5. Putting - B in place of B in (3), we get, , = cos2 A-sin2 8, , sin(A -B) =sin Acos B - cos AsinB, , (v), , 6. Putting - B in place of B in (4), we get, , tan(A-B) =, 7. cot(A+B), , tan A - tan B, , 1 + tan A tan B, , I Example 44. Find the value of tan 105°., -(vi), , Sol. tan 105° = tan(60? + 45°) = tan6QP * tan45°, 1 - tan60°tan45°, , 73 + 1 (73 + l)2, = -(2+73), 1-73-1, 1-3, tanl05° = -(2 + 73), , _cos(A + B), sin(A + B), , cos A cos B - sin Asin B, sin A cos B + cos A sin B, cot A cot B -1, , I Example 45. Prove that tan70° = tan20°+ 2 tan 50°., (vii), , cotB+cotA, , [dividing numerator and denominator by sin A sin B], 8. Putting - B in place of B in (7), we get, -cot AcotB-1, cpt(A-B) =, -cot B + cot A, cot A cot B +1, cot B - cot A, , (viii), , Third Proof by Complex Number Method, The result of the sine, cosine and tangent of compound, angle can also be derived using the concept of complex, numbers as discussed., , cos(A±B) + isin(A±B) -e, , i, , i(A±B), , Sol. tan70° = tan(20° + 50°) =, , or, or, , tan 20° + tan 50°, 1 - tan 20° tan 50°, , tan70° - tan20°-tan50°-tan70? = tan20° + tan50?, tan70° = tan70°tan50°tan50° + tan20° + tan50°, = cot 20° tan 50° tan 20? + tan 20? + tan 50°, [v tan70° = tan(90? - 20°) = cot20?], = 2 tan 50° + tan 20°, , I Example 46. If A + B = 45°, then show that, (1+tanA)(1+tanB) = 2., Sol. tan(A + B) =, , or, , = e“ ■ e/(±B) = (cos A + i sin A)(cos(± B) + i sin(± B)), , tan A + tan B, ■, j 1 —, 1-tan A tan B, , tan A + tan B, 1 - tan A tan B, , [as A + B = 45°, tan(A + B) = 1], tan A + tan B + tan A tan B = 1, 1 + tan A + tanB + tanAtanB = 1 + 1, [v adding T’ on both sides], (1 + tan A) + tanB(l + tan A) = 2, (1 + tanA)(l + tanB) = 2, , www.jeebooks.in, = (cosAcosB±icosAsinB + isinAcosB TsinAsinB), , = (cos A cos B + sin A sin B) + i (sin A cos A ± cos A sin B), , =>, =>

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www.jeebooks.in, 24, , Textbook of Trigonometry, , I Example 47. Find the value of, , tan 495°, cot 855°, , sin a cos P + cos a sin P _ sin (a + P), sin a cos P, sin a cos P, , Sol. tan 495° =tan (2.180° + 135°) = tan 135° = - 1, cot 855° = cot(4.180° + 135°), = cot 135° = - 1, ['/ cot(4.180° + 0) = cot 0], tan 495° Z1 = 1, cot 855° -1, Example 48. Evaluate sin • nn +(- l)n — •; where n is an, 4, integer., Sol. v sin(n + 0) = - sin0, , sin(nn+ 0) =(- 1)" sin 0 => sin] rm + (- 1)" —, 4, , i, , = (-!)' sin{(-1)"I, 4, = (- in-iysinl, 4, :. sin{(-l)"0} = (—l)"sin0, , [vsin(- 0) = -sin 0], , 1, = (-l)2n sin —= sin —=, 4, 4, !z, , 0, =0, sin a cos p, [vsin2(a + P) = 1 - cos2(a +P) = 1-1 = 0], , I Example 52. Prove that, sin(B-C) sin(C-A) + sin(A-B), cos B cos C cos Ceos A cos A cos 8, Sol. First term of L.H.S., . sin(B -C), cos B cos C, sin Bcos C, cos B cos C, , sin B cos C - cos B sin C, cos Bcos C, cos B sin C, cos B cos C, , = tan B - tan C, Similarly, second term of L.H.S. = tan C - tan A and 3rd, term of L.H.S. = tan A - tan B, Now L.H.S. = (tan B - tan C) + (tan C - tan A), + (tan A - tan B) = 0, , I Example 53. Show that tan 75° + cot 75° =4., , I Example 49. Prove that cos 18° - sin 18° = J2 sin 27°, , Sol. tan 75° = tan(45° + 30°) =, , Sol. R.H.S. = Ji sin 27° =Ji sin(45° - 18°), , tan 45° + tan 30°, 1 - tan 45° tan 30°, , 1+*, , = Ji(sin 45° cos 18° - cos 45° sin 18°), 1, , = Ji I -^= cos 18° —7= sin 18° |, , \Jz, , Jl, , J, and, , = cos 18° - sin 18°, = L.H.S., , ...(i), , *, , V3, i__ _Ji-i, cot 75° =, tan 75° Ji + 1, , .(ii), , Now, L.H.S. = tan 75° + cot 75°, , I Example 50. Show that cot f — + x\ cot f — - xl = 1, kA ), k44, ), n, 71, — + x cos ---- X, 4___ ,, Sol. LH.S. =---., |, n, sin I — + x, 2 71, , -2, , •, , 2, , .2, , sm---- sm x, 4, , 1, , (5/3-1) (5/3 + 1), , • 2, , — sm x, =1, - - sin2 x, 2, , I Example 51. If sin a sin p - cos a cos p +1 = 0, Prove, that 1 + cota tanP = O, Sol. Given, sin a sin P - cos a cos P + 1 = 0, cos a cos p -sin a sin P = 1, cos(a + P) = 1, Now, 1 + cos a tanP = 1 + C-°-S— • -S-*n ft, sin a cos P, , =>, , [from Eqs. (i) and (ii)], , _ (73 +1)2 +(5/3 — 1)*, , cos, , cos---- sm x, 4, , J3-1 + 5/3 + 1, , _ (4 + 25/3) + (4 - 25/3) 8, --------------------------------- = — =, 3-1- 2, , — K.n.o., , n sin a cos a, I Example 54. If tan p = ----------—. Prove that, 1 — n sin a, tan(a -p) = (1-n) tana., Sol. tan P = " Sin a c°s P, 1 - n sin2 a, n sin a cos a, cos2 a, 1, n sin2 a, cos2 a, cos2 a, , www.jeebooks.in, =>, , [dividing numerator and denominator by cos a]

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 25, , a, , n tan a, sec2 a - n tan2 a, n tan a, 1 + tan a - n tan a, n tan a, 1 + (1 - n) tan2 a, tan a - tan P, Now, L.H.S. = tan(a - p) =, 1 + tan a tan p, n tan a, tan a 1 +(1 - n) tan2 a, n tan a, 1 + tan a, 1 +(1 - n) tan2 a, , 1 Example 57. Let A, B, C be the three angles such that, A+B + C = n. If tan A-tanB = 2, then find the value of, cos A cos 8, cos C, , -.(i), , So/. Given,, , Let, , [v cos C = cos(n -(A +B)= - cos(A +B)], cos A • cos B, sin A sin B - cos A cos B, , [from Eqs. (i)], , tan a + (1 - n) tan3 a - n tan a, 1 + (1 - n) tan2 a + n tan2 a, , (1 - n) tan a +(1 - n) tan3 a, 1 + tan2 a, , (1 - n) tan a (1 + tan2 a), 1 + tan2 a, , tan A • tan B = 2, cos A cos B, cos A ■ cos B, y=, cos C, cos(A+B) ., , ----1-------- *- = !, tan A tan B - 1 2-1, ., ., , cos 10° +sin 10°, I Example 58. Prove that------------------ = tan 55°., r, cos 10° — sin10°, 1 + tan 10°, 1 - tan 10°, , . cos 10° +sin 10°, cos 10° - sin 10°, , tan 45° + tan 10°, 1 - tan 45° tan 10°, , Sol.---------------- =----------- =-------------------, , = tan(45° + 10°) = tan 55° (dividing by cos 10°), , = (1 - n) tan a, , 1, , I Example 55. Show that cos2 8 + cos2 (a+ 0), - 2 cos a cos 0 cos(a + 0) in independent of 0., Sol. cos2 0 + cos2(a + 0) - 2 cos a cos 0cos(a 4- 0), = cos2 0 + cos(a + 0) [cos(a + 0)- 2 cos a cos 0], = cos2 0 + cos(a + 0), , [cos a cos 0 - sin a sin 0 - 2 cos a cos 0], = cos2 0 - cos(a + 0) [cos a cos 0 +sin a sin 0], = cos2 0 - cos(a + 0) cos(a - 0), , Sol. tan 2A = tan[(A +B) + (A - B)], tan(A +B)+ tan(A -B), 1 - tan(A +B) tan(A -B), 7t, , = cos2 0 + sin2 0 - cos2 a, , Also,, , I Example 56. If 3 tan 0 tan 0 = 1, then prove that, 2cos(0 + 0) = cos(0 - 0)., Sol. Given, 3 tan 0 tan 0 = 1 or cot 0 cot 0 = 3, cos 0cos 0 3, or, sin 0 sin 0 1, , By componendo and dividend©, we get, cos 0 cos0 + sin 0sin <J) 3 +1, cos 0 cos 0 -sin 0 sin 0 3-1, or, or, , cos(0 ~ 4>), cos(0 + 0), , 2, , 2 cos(0 + 0) = cos(0 - 0), , It, , Given that, 0 < A < — and0 < B< —. Therefore,, 4, 4, , = cos2 0 - [cos2 a -sin2 0], = 1 - cos2 a, which is independent of0., , 2, , I Example 59. If sin(A -B) = -==, cos(A+B) = -7=,, VW, V29, find the value of tan 2A where A and B lie between 0, and —., 4, , 0< A + B< —, 2, n, n, ---- < A - B < — and sin( A - B) =, 4, , 4, , 0< A - B< —, 4, , Now,, , sin(A -B) = -4=, V10, , ...(ii), , tan(A - B) =, , A, , „, , 2, , cos(A + B) = - —-, , V29, , tan(A +B) = 2, From Eqs. (i), (ii) and (iii), we get, 5 1, -+—, 17 16 „, tan 2A = 2 3_.= —x —= 17, 5 1 6, 1, 1--X2 3, , ...(iii), , www.jeebooks.in

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www.jeebooks.in, 26, , Textbook of Trigonometry, , I Example 60. Prove that (1 + tan 1°) (1 + tan 2°)..., (1+ tan 45°) = 223., , _ tan 25° + tan 55° + tan 100°, tan 25° • tan 55° • tan 100°, , Sol. (1 + tan x°)(l + tan(45° - x0)), , Since,, 25° + 55° + 100° = 180°, tan 25° + tan 55° + tan 100° = tan 25° tan 55° tan 100°, =>, £=1, , .f, 1 - tan x°, = (1 + tan x' ) 1 +----------- - = 2, 1 + tan x°, , I Example 63. Prove that, , (1 + tan 1°)(1 + tan 44°), = (1 + tan2°)(l + tan 43°), = (1 + tan 3°)(l + tan 42°), , 1OO, , y sin(kx) cos(101 - k)x = 50 sin (101x), k=l, 100, , Sol. Let S = ^sin(fcx) cos (101 - k)x, k»l, , =(1 + tan 22°)(1 + tan 23°), =2, (1 + tan 1°)(1 + tan 2°)...(1 + tan 45°)= 2a, , S=sin x cos lOOx + sin 2x cos 99x, +... + sin lOOx cos x ...(i), 5 = cos x sin lOOx + cos 2x sin 99x +... +, sin x cos lOOx ...(ii), (on writing in reverse order), On adding Eqs. (i) and (ii), we get, 2S =(sin x cos lOOx + cos x sin lOOx), + (sin 2x cos 99 x + cos 2x sin 99 x), , =>, , (as 1 + tan 45° = 2), , I Example 61. If cos(P - y)+cos(y - a), + cos(a - P) = -1, Prove that, , cosa + cosP + cos y =sina+sinp + siny =0, 3, Sol. Given. cos(P - y) + cos(y - a) + cos(a - P) = —, 2, or 3 + 2 cos(P - y) + 2cos(y - a) + 2 cos(a - 0) = 0, , + (sin lOOx cos x +sin x cos lOOx), = sin lOlx +sin lOlx +... + sin lOlx (100 times), Hence, 5 = 50sin (lOlx), , or 3 + 2(cos 0 cos y + sin P sin y), , + 2(cos y cos a + sin y sin a), + 2(cos a cos p + sin a sin p) = 0, or(cos2 a +sin2 a) +(cos2 p + sin2 p) + (cos2 y + sin2 y), + 2(cos P cos y +sin P sin y) + 2(cos y cos a + sin y sin a), + 2(cos a cos p + sin a sin P) = 0, or(cos2 a +cos2 p + cos2 y + 2 cos a cos P + 2 cos P cos y, + 2 cos y cos y) + (sin2 a + sin2 p + sin2 y, , 7t, , I Example 64. If A = —, then find the value of, 8, , ^tan(M)-tan((r + 1)A)., r=1, , Sol. tan((r + 1)A - (rA)) =, , tan(r + 1)A- tan(rA), , 1 + tan(r + 1)A • tan(rA), , + 2 sin a sin p + 2 sin P sin y + 2 sin y sin a) = 0, or (cos a + cos p + cos y)2+ (sin a + sin P + sin y)2 = 0, , s, tan(rA) ■ tan(r + 1)A, , S=, r-1, , which is possible only when, cos a + cos p + cos y = 0 and sin a + sin P +sin y = 0, , =i(-D+ tan A /^(tan(r + 1)A - tan( rA)), r=l, , I Example 62. Find the value of, COt 55° +COt 100°, -I---------------------------, , tan 55° + tan 100°, , tan 25° + tan 55°, cot 100° +cot 25°, tan 100°+tan 25° ’, , = -8 + —— ■(tan 9 A - tan A), tan A, , Now,, , Sol,E= cot 2 5°J cot.55° + cot 55° + cot 100°, tan 25° + tan 55° tan 55° + tan 100°, , ♦------ 1------ ------- 1____, tan 55° tan 100°, , tan 100° tan 25°, , 5, , 7t, = - tan —, 5, , tan 100° + tan 25°, tan 55° tan 100°, , 971, tan 9A = tan —, 5, f„, 71, = tan 27t----, , I, , + cot 100° + cot 25°, , 1, , /■i, , COS 25° + COt 55°, , =>, , S = -8 + —— (- 2 tan A), tan A, , www.jeebooks.in, = -8-2 = -10

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 27, , I Example 65. Prove that, sin 0-sec 30 + sin30-sec 320 + sin32 0-sec 33 0 + ..., , 4 4, , 3 3, , 5 5, , 5 5, [using cos(a + P) = 3 / 5, cos(a - P) = 4 / 5, , =---- +----, , upto n terms = [tan 3n 0 - tan 0], , =>sin(a + P) = 4 /5, sin(a - P) = 3/5], , 16 + 9, , Sol. sin 9 • sec 39 + sin 39 • sec 32 9 + sin 329 • sec 33 9 +... upto n, , 25, , terms, , = 1 => sin 2a = 1, , n, , /, , = ^sin3r-1 9 sec3f 9, , I Example 68. If cosa =, , y 2 cos 3r-1 9 sin 3r 1 9, “f 2 cos 3'"‘ 9 • cos 3' 9, , =-X, 2~ cos 3r-1 9 • cos 3r 9, _ly sin(3r9 - 3r'‘9), cos 3'-1 9 • cos 3r9, , ”2, , sin 3r 9 • cos 3r“‘ 9, , =-i, , then evaluate cos(a - P)., i, cos a = -| x + —1, x, 2, , Sol., , sin(2-3r“‘ 9), , - cos 3r 9 - sin 3r‘‘ 9, , cos 3r-‘ 9 • cos 3r 9, , 1, n, ft 1if 1), x + — L cosp = — y+ - ,, x, 22^, yj, , => x2 - 2xcosa +1 = 0 => x, , 2cosa ± ^/4cos2a - 4, 2, , 2cosa ± 2isina, =>, x --------------------2, x = cosa ± isina, Similarly, y = cos P ± i sin P, x cosa ± isina, , {as V-l = i}, , = cos(a - p) ± isin(a - 3), , y, , = - ^(tan 3r 9 - tan 3r"‘ 9), 2 r=l, , example 66. In a triangle ABC, if, sm Asin(B-C)=sinCsin(A-B), then prove that cot A,, cot 8, cotC are in AP., , x, , sin(A -B), , sin C sin B, , sin A sin B, , sin B cos C -sin C cos B, , sin C sin B, , sin A cos B - sin B cos A •, sin A sin B, , cot C - cot B = cot B - cot A, 2 cot B =cot A + cot C, :. cot A, cot B, cot C are in A.P., , cosa ± isina, , On adding Eqs. (i) and (ii), we get, — + — = 2cos(a - P), y x, , cos(a - p) = -| — + — ., 2<y x), , i.e., , Sol. sin Asin(B -C) = sin C sin(A -B), , sin(B - C), , —(i), , y = cosP ± ismp =, , and, = - [tan 3"9 - tan 9], 2, , cosp±isinP, , I Example 69. If 2sinacos0siny = sinpsin(a + y)., Then, show tana, tan 3 and tany are in harmonic, progression., Sol. We have, 2sina cosPsiny = sinpsin(a + y), , or 2sina cosPsiny = sinp{sina cosy + cosasiny}, => 2sinacosPsiny = sinasinpcosy + cosasinPsiny, On dividing both sides by sin a sin P sin y, we get, , I Example 67. If 0 < 3 < a < n /4, cos(a + 0) = 3/5 and, cos(a — 3) = 4 /5, then evaluate sin2a., , „, , Sol. We know, sin(2a) = sin{(a + P) + (a - p)}, , i.e., , 111, ., ------ ,------ ,------ are in AP, tana tanP tany, , or, , tan a, tan P, tan y are in HP., , O, , 2cotp = cota + coty, , = sin(a + p) • cos(a - P) + sin(a - P) • cos(a + P)., , 2, , or ---- —, , tanP, , 1, , 1, , ------ +------tana tany, , www.jeebooks.in

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www.jeebooks.in, 28, , Textbook of Trigonometry, , Exercise for Session 6, , 1. If a lies in II quadrant, 0 lies in III quadrant and tan (a + 0) > 0, then (a + 0) lies in, , quadrants., , 2. If 3 tan A tan B = 1, then prove that -°- ——— = 2., cos (A + B), , 3. If tan a =, , m, __ 1, and tan 0 =, the find the value of a + 0., m +1, 2m + 1’, , 4. lfcos(a + 0) =, , i, 71 i, 4, 5, sin(a-0) = — and a, 0 e 10, — I, then find the value of tan 2a., 5’''' 13, , 5. If a + 0 = — and 0 + y = a, then find the value of tan a., , 2, , -, , ', , -, , 6. If cos(0 - a) = a and cos(9 - 0) = b then the value of sin2 (a - 0) + 2a b cos(a - 0)., , 7. If 2 cos A = x + —, 2 cos 0 = y + — then show that 2 cos(A -8) = — + —., x, y, y x, 8. If y =(1+ tan A)(1- tan8), where A-B = —, then find the value of(y + 1)y + 1., 4, , Session 7, Sum of Sines/Cosines in Terms of Products, , Converting Product into Sum/, Difference and Vice-Versa, , Above four formulas are used to convert product of two sines, and cosines into the sum or difference of two sines and cosines., , Sum/Difference into Products, , Product into Sum/Difference, , 1. sin A + sin B =2 sin, , 1. 2 sin A cos B = sin( A+ B)+sin( A-B), , cos, , A-B, , 2, , 2. 2 cos A sin B = sin( A + B) - sin( A - B), , 3. 2 cos A cos B =cos(A + B) +cos(A - B), , 2. sin A-sinB =2 cos, , 4. 2 sin A sin B = cos( A - B) - cos( A + B), Proof We know that, sin A cos B + cos A sin B =sin(A + B), sin A cos B - cos A sin B = sin(A -B), cos A cos B - sin A sin B = cos( A + B), cos A cos B + sinA sin B =cos(A - B), , A+B, , 3. cosA+cosB=2cos, , ...(i), -(ii), -(iii), -(iv), , Adding Eqs. (i) and (ii), we obtain, 2sinAcos B =sin(A + B)+sin(A -B), , (v), , Subtracting Eqs. (ii) from (i), we get, 2 cos A sinB ~sin(A + B) -sin(A - B), , ,(vi), , 4. cos A - cos B = 2 sin, , 2, , A + Ba . f A-B, 2 7 5U\—, A-B^, , 2, , 7, A + B^, 2, , •cos, •sin, , 7, , A+B, 2, , B-A, 2, , Proof (i) Let A = C + D and B =C -D, then C =, , A+B, 2, , and, , Adding Eqs. (iii) and (iv), we get, 2 cos A cos B = cos(A +B) + cos(A -B), , ...(vii), , Subtracting Eqs. (iii) from (iv), we get, 2 sin A sin B = cos( A - B) - cos(A + B), , ...(viii), , 2, , L.H.S. = sin(C + D) +sin(C -D) = 2 sin Ceos D, A+B, A-B, = 2 sin ------- cos-------- =R.H.S, 2, 2, , www.jeebooks.in, Similarly we proof of (ii), (iii) and (iv).

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 29, , I Example 73. If sin A=sin 8 and cos A = cos B, then, A-B, prove that sin ------ =0., 2, , Some other Useful Results, , , ., x ., r, sin(A + B), ,, ., n, 1. tan A + tan B = —---------- , where A, B / mt + —, cos A cos B, 2, , Sol. We have sin A = sin B and cos A = cos B, , » x, . ., sin(A-B), ,, x, n, 2. tan A - tan B = —-------- where A, B # mt + —, cosAcosB, ., 2, , or, , sin A - sin B = 0 and cos A - cos B = 0, A -B, A +B, 2 sin, sin, =0, 2, 2, , x ., , t,, , sin(A + B), , ., , x „, , or, , , x, , x n, , sin(B -A), , ., , x, , and -2 sin, , 3. cot A + cot B = — -------- where A,B*mt,ne z, sin Asin B, 4. cot A - cot B =-------------where A, B * mt, n G z, sin Asin B, , or sin, , I Example 70. Prove that, cos 55° + cos 65° + cos 175° = 0., , A-B, A + BV, sin, =0, 2 J, 2, , A -B, , = 0, which is common for both the equations., , 2, , I Example 74. Prove that sin 20° sin 40° sin 80° = y-, , Sol. L.H.S. = cos 55° + cos 65° + cos 175°, , 55° + 65°, , 55° - 65°, , 2, , 2, , Sol. LH.S. = sin 20° sin 40° sin 80° = - (2 sin 80° sin 40°) sin 20°, 2, , = 2 cos ----------- cos-------------+ cos 175°, = 2 cos 60° cos(-5°) + cos 175°, , = - [cos(80° - 40°) - cos(80° + 40?) sin 20°, 2, , = 2 X - cos 5° + cos(180° - 5°), 2, , (cos 40° - cos 120°) sin 20°, , =, , = cos 5° - cos 5° = 0, , = - (2 cos 40° sin20° - 2 cos 120° sin20°), 4, , I Example 71. Prove that, sin A + sin 2A + sin 4A + sin 5A, ------------------------------------- = tan 3A., cos A + cos 2A + cos 4 A + cos 5A, , = | [sin(40° + 20°) -sin(40° - 20°)- 2, , [sin 60° - sin 20° + sin 20° ]= — sin 60°, , =, , sin A + sin2A + sin 4A + sin5A, cos A + cos 2A + cos 4A + cos 5A, , (sin5A + sin A) + (sin 4A + sin2A), , 4, , (cos 5A + cos A) + (cos 4A + cos 2A), , 2 cos 3A cos 2A + 2 cos 3A cos A, 2 cos 3A(cos 2A + cos A), , 2, , a-P, 2, , a-p, , sin 3A, , = ” sin A[cos(60° + A - 60° + A), - cos(60° + A+60° - A)], , 2, , = ~ sin A(cos 2A - cos 120°), , 2, , + <2 sin, = 4 cos2, , 8, , = ~sin A[2sin(60° +A)-sin(60° - A)], , Sol. L.H.S. = (cos a + cos 3)2 + (sin a + sin P)2, , a-P, , 2, , Sol. L.H.S. = sin A -sin(60° - A) -sin(60° + A), , I Example 72. Prove that (cos a + cos 3)2, a -PA, + (sin ex + sin P)2 =4 cos2, 2 J, , cos, , J3, , • sin(60° + A) =, , 2sin3A(cos 2A + cos A), , a +3, , J3, , I Example 75. Prove that sin A-sin(60° - A), , 2 sin3A cos 2A + 2 sin 3A cos A, , = • 2 cos, , 1 sin 20° ], 2, , cos, , a +3, , cos, , 2, , 2«+3, 2«+P, + sin, 2, 2, , a -P, , = “ (2 cos 2A sin A - 2 cos 120° sin A), , 2, = — sin(2A + A) - sin(2A — A) — 2| —, , 4, , I, , = — (sin 3A - sin A - sin A) = — sin 3A, 4, 4, , sin A, , www.jeebooks.in, = 4 cos2, , 2, , = R.H.S.

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www.jeebooks.in, 30, , Textbook of Trigonometry, , Exercise for Session 1, , 1. Show that sin x + sin 3x + sin 5x + sin 7x = 4 sin 4x cos 2x cos x., 2. Show that sin A • sin(8 - C) + sin 8 • sin(C - d) + sin C ■ sin(d - 8) = 0., , a+0, 0+ y, Y+a, 3. Show that cos a + cos 0 + cos y + cos(a + 0 + y)=4 cos ----- - • cos -—- • cos, 2, 2, 2, 3, 3, 4. If x and y are acute angles, such that cos x + cos y = - and sin x + sin y = -, then the value of sin(x + y)., 2, , n, , 9n, , 3n, , 5n, , 5. Find the value of expression 2 cos — cos — + cos — + cos —., 3, 13, 13, 13, 6. Find the value of, , cos d + cos 8, K sin A -sin 8, , 7. Find the value of 11 + cos -, , I, , 8., , n, , +, , sin A +sin8, cos A - cos B, , n, , (where, n is an even), , ., 3n, 57tY„, 7n, 7nA, 1 + cos — 1 + cos — 1 + cos — ., 8., 8, 8, , 8. In a triangle ABC, cos 3d + cos 38 + cos 3C = 1, then find any one angle., , Session 8, Trigonometric Ratios of Multiples of an Angle, Trigonometric Ratios of, Multiples of an Angle, Definition An angle of the form nA, where n is an integer, is called a multiple angle, for example 2A, 3A, 4A,... etc., are multiple angles of A, , In this session we shall express trigonometrical ratios, of multiple angles of A in terms of trigonometrical ratios, of A, , Trigonometrical Ratios of 2A in term, of Trigonometrical Ratio of A, ., , . „ 4, , „ ., , x, , 4, , 2 tan A, , 1. sin 2 A = 2 sm A cos A =------------1 + tan2 A, , 2. cos2A =cos2 A-sin2 A = 1 -2sin2 A, 2 4-1“ tan2 A, , = 2 cos A -1 =---------—, 1 + tan2 A, , 3. 1 +cos2A =2cos2 A, 1 -cos2A = 2sin2 A, , 77, 2 tan xA., 4. tan 2A = ----------where, A, #=, (2n, +1), —, l-tan2 2A, 4, , Proof sin 2 A = sin( A + A) = sin A cos A + cos A sin A, [using the formula sin( A +B) =sin A cos B + cos Asin B], , = 2sinAcos A, cos 2 A = cos(A + A) =cos A cos A - sin A sin A, , = cos2 A - sin2 A, tan 2A = tan (A + A), tan A + tan A, , 1 - tan A tan A, , 2 tan A, 1 - tan2 A, , Trigonometrical Ratios of 34 in terms, of Trigonometrical Ratio of A, 1. sin3A = 3sinA-4sin3 A, , = 4 sin(60° - A) • sin A • sin(60° + A), 2. cos3A = 4cos3 A-3cosA, , = 4 cos(60° - A) cos A cos(60° + A), , www.jeebooks.in, l + cos2A, 2 . l-cos2A, cos2 A, A,,------------- = sin2 A, or------------ ==cos, 2, 2, , 3. tan 3d =, , 3 tan A - tan3 A, 1 - 3 tan2 A

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , Proof, , 3, v tan A = 4, , 1. sin 3 A = sin(2 A + A) = sin 2A cos A + cos 2A sin A, , = 2 sin A cos A-cosA+(l-2sin2 A) sin A, , sin 4A = sin 2A cos 2A = 2 x — x — = —., 25 25 625, • 24, V sin 2A = —, 25, , = 2 sin A cos2 A+sinA-2sin3 A, , 1, , = 2 sin A(1 - sin2 A) + sin A -2 sin3 A, , 7, and cos 2A = —, L, 25j, , = 2 sin A - 2 sin3 A + sin A - 2 sin3 A, , 1, , = 3 sin A - 4 sin3 A, 2. cos 3A = cos(2A + A) = cos 2A • A cos A - sin 2A sin A, , ■, , 31, , § Example 77. Prove that, , = (2cos2 A -l)cos A-2 sin A cos A - sin A, = 2 cos3 A-cos A-2cos A(l-cos2 A), , Sol. L.H.S. = -*, 1 -sin 20, , = 2 cos3 A - cos A - 2 cos A + 2 cos3 A, , sin2 0 + cos2 0 + 2 sin 0 cos 0, sin2 0 + cos2 0 - 2 sin 0 cos 0, , sin 0 + cos 0, , 1 + tan 0, , sin 0 - cos 0 /, , 1 - tan 0, , 3 sin A - 4 sin3 A, , 2, , [dividing numerator and denominator by cos 0], , 4 cos3 A - 3 cos A, , sin A(3 - 4 sin2 A), , tan A(3 - 4sin2 A), , cos A(4cos2 A-3), , 4cos2 A-3, , On dividing by cos2 A numerator and denominator, , tan A(3 sec2 A - 4 tan2 A), , 1- tan2 2-4, 4, * Example 78. Prove that------1+ tan2, , U, , Sol.--------- _4____ ,, , 1 - tan2 0, , i-A, , 1 + tan2 0, , 1 + tan2, , tan A(3 + 3 tan2 A - 4 tan2 A), 4-3-3 tan2 A, tan A(3 - tan2 A), , 3 tan A - tan3 A, , 1-3 tan2 A, , 1-3 tan2 A, , 1 Example 79. Prove that, Sol. We have, LHS =, , =>, , LHS =, , 25, , 5, , =>, , sec 40 -1, , sec 40-1, , 1 1, ---cos 80, , LHS=2sini 48, cos 88, , 4, , sin 2A = 2 sin A cos A = 2 x - x - = —, 5 5 25, , -V5 J, , tan 80, tan 20, , 1 - cos 80, , cos 40, , cos 80, , 1 - cos 0, , cos 40, , 3, , cos 2A = 1 - 2 sin2 A = 1 - 2 x, , sec 80 -1, , sec 80 - 1, , —1---- 1, 4, , ), , U, , 3, Sol. We have, sin A = -, where 0° < A < 90°, 5, :., cos2 A = 1-sin2 A, , '1-1, , where — - A =0, 4, , = cos 20 = cos — - 2A = sin 2A, , 3, where 0° < A < 90°, find the, 5, values of sin 2A, cos 2A, tan 2A and sin 4A., , sin A, tan A =------cos A, , J, , .4, , I Example 76. If sin A=, , - sin2 A =, , =sin2A., , 1 + tan2|, , 4 - 3 sec2 A, , cos A = +, , 1-sin 20, , 2, , = 4 cos3 A-3 cos A, , t, . sin3A, 3. tan 3A =--------cos 3 A, , ' 1 + tan 0 \2, k 1 - tan 0}, , 1 +sin 20, , 7_, , 25, , =>, , 2x1, , 6, , 4, , 4, , 24, , 1-1, , 7, , LHS =, , cos 40, 2 sin2 20, , 80, v 1 - cos 80 = 2 sin2 — = 2 sin2 40, 2, 40, and, 1 - cos 40 = 2 sin2 — = 2 sin2 20, 2, (2 sin 40 cos 40), sin 40, cos 80, , X------- —, , 2 sin* 20, , www.jeebooks.in, tan 2A =, , 2 tan A, , 1 - tan2 A, , 1-, , -T4J, , 16, , LHS =, , 2 sin 40 cos 40 |f 2 sin 20 cos 20, cos 80, , x, , 2 sin2 20

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www.jeebooks.in, 32, , Textbook of Trigonometry, , cos 26", f sin 2(46), x, I cos 86, sin 28 >, /, sin 88, cos 26, x, = tan 86 cot 26, ^sin 26, cos 86, , LHS, , 7t, 8, , 8, , and, , 5n, 8, , 3n, , 37C, , 8, , 8, , 371, 8, , LHS = 1 + cos —, , = RHS, , I, , tan 26, , it, , + ^2 + ^/2 + 2cos8 = 2 cos 0, , LHS = 72 + 72 + 72(2 cosZ2 40), 80, , => LHS = —, 4, , 4, , 1 - cos 6=2 sin, , 1-4=, , => LHS =—, 4, , LHS = 72+72 +2 cos 40, LHS = 72 +72(1 + cos 40), LHS = ^2+72 (2 cos2 26), , [v 1 + cos 40 =2cos226], , LHS =72 + 2 cos 20 =5/2(1 +cos 26), , ,0', , V2., , I Example 81. Show that 5/3 cosec 20° - sec 20° = 4, Sol. We have, LHS =5/3 cosec 20° - sec 20°, , cos 28 =, , LHS =, LHS =, , 1, , Sol. We have,, , 2< — cos 20° — sin 20°, 2, 2, , - 2tan2 0, 2+ 2 tan2 0, , - tan2 0, sec2 0, , cot 6 - tan 6 =, , =* LHS =, , 4 sin 40°, , - sin2 <|), , =>, , 2 sin 40°, sin 20° cos 20°, , sin 40°, , 71, , (, , I Example 82. Prove that 11 + cos — 1 + cos —, , 2, , - tan 6, , tan 6, , 4 s'n 4-°^ = 4 = RHS, , 2 sin 20° cos 20°, , 1, tan 6, , 1 - tan2 6, , sin 20° cos 20°, , 7n, 8, , [v tan2 8=2 tan2 <f) + 1], , 1 + 2 tan2 0 + 1, , 2(sin 60° cos 20° - cos 60° sin 20°), , sin 20° cos 20°, , 571, , 1 - (2 tan2 0 + 1), , sin 20° cos 20°, , 2 sin(60° - 20°), , - = RHS, 8, , tan a + 2 tan 2a + 4 tan 4a + 8 cos 8a = cot a, , sin 20° cos 20°, , Lrlo =------------------------, , 2, , I Example 84. Prove that, , cos 20° - sin 20°, , *, , 1-1, , cos 20 + sin2 0 = 0, , LHS = ——---------- -—, sin20° cos 20°, , V3, , 4, , 1 Example 83. If tan2 0 = 2 tan2 0+1, prove that, cos 20 +sin2 0 = 0., , cos 26 =, , LHS =, , 1, , 1 + -|V2., , 2, , Sol. We have, cos 20 =-—tan, 1 + tan2 6, , = 72(2cos2 8) = 2 cos 6 = RHS, , =*, , 37t, , 1 - cos---, , 2, 1, , =>, , 2 3tc, , LHS = ^2 + 72 + 7(4 cos2"46), , =>, , =>, , 2 37C, 8, , 1 - cos —, , 7t, 1 - cos —, 4, , 1, , v 1 + cos 86. = 2 cos2, , =>, , 2 7t, 8, , 71, 1 - cos —, 8, , => LHS = sin2 — sin, 8, 8, 2 37t, 1, 2 7t, 2 sin, => LHS =— 3 sin, 8, 8, 4I, , Sol. We have, LHS = 72 + 72+ 72 (1+cos 86), , =>, , .1 - cos —, 371, 8, , 1 + cos---, , 8, , => LHS = 11 — cos, , I Example 80. Show that, , = - COS —, , cos — = cos 71------ = - cos —, 7t, , LHS =, , 7t, , 7lt, 8, , cos — = cos It----, , 8, , cot 6 - tan 8 =, , =2, , 1 - tan2 6, , 2 tan 6, , 2, , tan 26, , =>, , cot 6 - tan 6 = 2 cot 26, (i), We have to prove that, tan a + 2 tan 2a + 4 tan 4a + 8 cot 8a = cot a, cot a - tan a - 2 tan a - 4 tan 4a - 8 cot 8a = 0, or,, . Now,, LHS = cot a - tan a - 2 tan 2a - 4 tan 4a - 8 cot 8a, =>' LHS = (cot a - tan a) - 2 tan 2a - 4 tan 4a - 8 cot 8a, => LHS = 2 cot 2a - 2 tan 2a - 4 tan 4a - 8 cot 8a, [using (i)], => LHS = 2(cot 2a - tan 2a) - 4 tan 4a - 8 cot 8a, , www.jeebooks.in, I, , 8, , Sol. We have,, , 1 + cos —, , 8’

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www.jeebooks.in, Chap 01, , =>, =>, , => LHS = 2(2 cot 4a) - 4 tan 4a - 8 cot 8a, [On replacing 0 by 2a in Eq. (i)], => LHS = 4 cot 4a - 4 tan 4a - 8 cot 8a, => LHS = 4(cot 4a - tan 4a) - 8 cot 8a, => LHS = 4(2 cot 8a) - 8 cot 8a, [On replacing 0 by 4a in Eq. (i)], => LHS = 8 cot 8a - 8 cot 8a => LHS = 0 = RHS, , I Example 85. Determine the smallest positive value of, x (in degrees) for which tan(x +100°) = tan(x + 50°), tan x tan(x - 50°), Sol. We have, tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°), tan(x + 100°), =>, = tan(x + 50°) tanx°, tan(x + 50°), sin(x + 100°) cos(x - 50°), cos(x + 100°) sin(x - 50°), , =>, =>, , Trigonometric Functions and Identities, , 33, , sin(2x + 50°) cos(2x + 50°) = - sin 150° cos 50°, 2 sin(2x + 50°) cos(2x + 50° )= - 2 cos 60° cos 50°, [’.'sin 150° = cos 60°], sin(4x + 100°) = sin(270 - 50°), sin(4x + 100°) = sin 220°, 4x + 100° = 220° => x=30°, , I Example 86. Prove that, sinx, sin3x, sin9x, 1., -------- +--------- +----------- = - (tan 27 x - tan x), cos 3x cos 9x cos 27 x 2, sin 3x, sin 9x, Sol. We have, --- - ■ +-------- +, cos 3x cos 9x cos 27 x, , sin(x + 50°)sin x, cos(x + 50°) cos x, , 1 [ 2 sin x cos x + 2 sin 3x cos 3x, ” 2, , 1, , sin(x + 100°) cos(x - 50°) -cos(x + 100°) sin(x - 50°), sin(x + 50°)sin x + cos(x + 50°) cos x, , cos3xcosx, sin 2x, , 2 cos 3x cos x, , sin(x + 100°) cos(x - 50°)+ cos(x + 100°)sin(x - 50°), , 2 sin 9x cos 9x, +-----------------cos9xcos3x cos 27 x cos 9x, , sin 6x, sin 18x, +------------------ F, cos 9x cos 3x cos 27 x cos 9x, , sin(3x - x), , sin(9x-3x), , sin(27x-9x), , 2 cos3xcosx, , cos9xcos3x, , cos27xcos9x, , _1, , ---- - ------------------- 4"---------------- ------ F-----------------------, , sin(x + 50°) sin x - cos(x + 50°) cos x, =>, , =>, , sin(x + 100° + x - 50°), , cos(x + 50°-x), , sin(x + 100° - x + 50°), , - cos(x + 50° + x), , sin(2x + 50°), , cos 50°, , sin 150°, , - cos(2x + 50°), , =, , {(tan 3x - tan x) + (tan 9x - tan 3x), + (tan 27x - tan 9x)}, , = - (tan 27 x - tan x), 2, , Exercise for Session 8, 1. This question has statement which is true or false., 7t, , 71, , t, , If - < 0 < —, then the value of ^1 - sin 2 0 = cos 0 - sin 0., , 1 - COS 2 0, 3tc, 2. If n < 0 < —, then find the value of, 1 +cos 20", 2, , o, 4, x, 3. If tan x = - -, x lies in II quadrant, then find the value of sin —., 2, , 4. Prove that sin4 — + sin4 — + sin4 — + sin4 — = -., 8, 8, 8, 8 2, 5. If A = 2 sin2 0 - cos 2 0 and A e [a, 0], then find the values of a and 0., , 1, 6. Ifsinx + cosx = -, then find the value of tan 2x., 5, , 7. If tan 34 =, , 3 tan 4 + k tan3 4, -, then k is equal to, 1 - 3 tan2 4, , 8. If tan A + 2 tan 2 /I + 4 tan AA + 8 cot 8A=k cot A then find the value of k., , www.jeebooks.in, 8n, 4tc, 27t, 14n = n, _22, then find the ...., 9. If m2 cos — cos — cos — cos —, value of, 15, 15, 15 " ’, 15, , 2 -n, _n2, , n2, , 10. If(2n + 1) 0 = 7t, then find the value of2n cos 0cos 20cos 220...cos 2n -10.

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www.jeebooks.in, Session 9, =, , i, , =, , Trigonometric Ratios of Submultiple of an Angle, , ■, , !, , Definition, An angle of the form —, where n is an integer is called a, n, submultiple angle of A., AAA, For example, etc., are submultiple angles of A., 2 3 4, , In this session we shall express the trigonometric ratios of, A in terms of the trigonometric ratios of submultiple, A A, angles, etc., and vice-versa., 2 3, , Trigonometric Ratios of A in Terms of, Trigonometric Ratios of 2, A, (i) sin 2 A = 2 sin A cos A Putting — in place of A, we get, 2, sin A = 2 sin — cos —, 2, 2, , (ii) cos 2A =cos2 A - sin2 A Putting — in place of A we, ., ., 2 A . 2 A, get cos A=cos-----sin —, 2, 2, A, (iii) cos 2 A = 2 cos2 A -1 Putting — in place of A, we get, 2, 7 A, cos A = 2 cos-----1, 2, , A, (iv) cos 2 A = 1 - 2 sin2 A. Putting — in place of A, we get, 2, 2 A, cos A = 1 - 2 sin, 2, 2 tan A, (v) tan 2A =, 1 - tan2 A, 2 tan —, 4, .’. tanA =-------- -—, putting — in place of A, 1-tan2 2, , (vi) sin 2 A =, , 2 tan A, 1 + tan2 A, , 2 tan —, A, •, sin, AA =-------- 2-—, putting —, 2, 1 + tan2 —, 2, , in place of A, , ...., n a 1-tan2 A, (vn) cos 2A =------------1 + tan2 A, , i, , j, , 2 A, 1-tan, A, cos A —-------- 2 putting —, 2 A*, 1 + tan, 2, , in place of A, (viii) cot 2 A =, , —1, 2 cot A, , cot2 — -1, 9, A, cot A =------ ------, putting —, 2 cot —, 2, 2, , in place of A, , Trigonometric Ratios of A in Terms of, A, Trigonometric Ratios of (i) sin 3 A = 3 sin A - 4 sin3 A. Putting — in place of A, we, , get sin A = 3 sin — - 4 sin3 —, 3, 3, (ii) cos3A = 4cos3 A-3cos A, .. cos A = 4 cos — - 3 cos, , putting — in place of A, , 3 tan A-tan3 A, . A. ,, _ 4, (iii) tan 3 A = ------------- ——, putting — m place of A, , A, 3 tan---- tan 3, :. tan A =------- --------- 3, A, 1-3 tan2, 3, , A . A, A., Values of cos sin - and tan - in, 2, 2, 2, Terms of cos 4, , www.jeebooks.in, ..., 2 A 1 + cos A, (i) cos — =----------2, 2, , A, cos — = ±, 2, , 1 +cos A, 2

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www.jeebooks.in, -3, , Chap 01 Trigonometric Functions and Identities 35, , A, sin — = ±, 2, , 1 - cos A, , 1 - cos A, A, (iii) tan2—= ---------- .'. tan — = +, 2 1 + cos A, 2, , 1 - cos A, , (ii) sin2 - =, 2, , 1-cos A, , 2, , Values of Trigonometric Ratios of, Some Particular Angles, , 2, , I. (i) Value of sin 18°, , 1 +cos A, , Let 0 = 18°, then 50 = 90°, , sin 20 = sin(90° - 30), , or 20 = 90° - 30, , Note, , A, , A, , 20 + 30 = 90°, , If cos A is given, then there will be two values of cos -.sin - and, , or sin20 =cos30 or 2sin0cos0 =4cos3 0 -3cos0, , A, A, tan - but if A is given, then there will be only one value of cos-., , or 2sin0 = 4cos2 0 -3, , A, A, sin - and tan - because + sign or - sign before the radical sign, , or 2 sin 0 = 4(1 - sin2 0) - 3 = 1 - 4 sin2 0, , [dividing by cos 0], , or 4 sin2 0 + 2 sin 0 -1 = 0, , A, can be fixed by knowing the quadrant in which - lies., , sin 9 =, , -2+27? -1 + 7?, , 2, , 8, , 8, , A, A, Values of sin - and cos - in Terms of, 2, 2, , Thus sin 0 =, , sin A, , v 0 = 18°, , 4, , -1 + 7? -1-7?, 4, , ’, , 4, , sin 0 = sin 18° > 0, for 18° lies in the 1st quadrant, , A . A>, A, 2 A . 2 A, A, cos — + sin — = cos — +sin — + 2cos — sin—, 2, 2, 2, 2, 2, 2J, , sin 9 i.e.,sin 18° =, , 7?-l, 4, , = 1 + sin A, (ii) Value of cos 18°, , cos — + sin — = ± 71 + sin A, 2, 2, , /, , cos — - sin — = ± 71-sin A, 2, 2, , Similarly,, , = 1-, , cos — = ± - 71 + sin A ± - 71 ~ sin A, 2, 2, 2 ', , ...(iii), , 2, , A, A, if A is given, there will be one and only one value of cos- and sin—, , because the + or - sign can be fixed before the radical sign in the, following way, 1 . /tT, 1cos -A, cos + sin - = 72 I-7, , .72, 2 + -7=sm72, 2, .2, 2j, = 72 (sin-cos - +• cos - sin-^ 72 sin, I 4, 2, 4 2, 4, , <4, , /, , 6-27?, 16, , _ 16-6+27? _ 10+ 27?, 16, [v cos 18° >0], , (iii) Value of tan 18°, , A, A, If sin A is given, then there will be 4 values of sin - and cos - but, , A, Ar, Similarly, cos - - sin - = 72 cos;(*, , 16, , = 1-, , V, , cos 18° = - 710+27?, , Note, , 2, , 5+1-27?, , 16, , Subtracting Eq. (ii) from Eq. (i), we get, sin — = ± - 71 + sin A + -1 71- sin A, 2, , 4, , ...(ii), , Adding Eqs. (i) and (ii), we get, , 2, , 5-1, , cos2 18° = 1-sin2 18° = 1-, , ,o0 sin 18° 7?-l, tan 18° =-------- =-------- +, cos 18°, 4, , 710+27?, 4, , 7?-i, 710+27?, (iv) Value of cos 72° and sin 72°, , (a) cos 72° = cos(90° -18°) =sin 18° =, , 2., , + 2?, 2., , 75-1, 4, , (b) sin 72° = sin(90° -18°) = cos 18° = - ^0+2^5, 4, , www.jeebooks.in, A, A, A, A, Thus the sign of cos - + sin - and cos - - sin - can be fixed by, 2.2, 22, 2, 71, /A, knowing the quadrant in which - + - lies., , n. (i) Value of cos 36°, , \2, , cos 36° = 1-2 sin218° = l-2x, , 7

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www.jeebooks.in, 36, , Textbook of Trigonometry, , = l-2x, , 5 + 1-2V5, , 1°, (ii) Value of cot 82 —, 2, , 16, , 1°, (, 11°>, cot 82 — = cot| 90° - 7 —, 2, 2>, , 3- 75, , = 1-, , 4, , = tan 7—=(73 - 72)(7F-1), 4, , 2, , 4, , (iii) Value of cot 7 —, 2, 1°, Let0=7—, then20 = 15°, 2, , Thus, cos 36° =, , 4, (ii) Value of sin 36°, \2, , sin22 36° = 1 -cos22 36° = 1-, , _1 6 + 275, , .1 4 J, 10-2^5, , 16, , 16, , sin 36° = -710-275, , .n, , 2-Ji _ 272+73+1, 73-1, y^-i, 272, (272 + 7i +1) ("73^ +1), (73 - 1) (^ +1), , (iii) Values of sin 54° and cos 54°, (a) sin 54° = sin(90° - 36°) =cos 36° =, , 4, , 2 Tb + 3 + 73 + 2^2 + y/3 +1, , (b) cos 54° = cos(90° -36°) = sin 36° = - (^10-275), 4, , 3-1, , = 2^+2^f2^4-4=^ + ^+^+2, , HI. (i) Value of tan 7 —, 2, , Let, , Now,, , ■, , 1+, , [vsin 36° >0], , 4, , 14-cos 15°, sin 15°, , 1+COS 20, , Now, cot 0 =-----------sin 20, , 16-6-275, , 16, , .*., , xt, , 2, , 0=7—.then 20 = 15°, 2, n 1 - cos 20, tan0 =-----------sin 20, [’.• 1 - cos 20 = 2 sin2 0 and sin 20 = 2 sin 0 cos 0], , 1-cos 15°, , 1-, , 2V2, , 7i-1, 2V2, 2^2 — 7s — 1 (272 - 73 -1) (7i +1), sin 15°, , (73 -1) (73 +1), , 73-1, , 2^6 -3-73+2a/2-a/3 -1, 3-1, , = 73(72+1)+72(72+1), = (73 + 72) (^ + 1), cot 7 — = (73 + ^2) (72 +1), 2, , 1°, (iv) Value of tan 82 —, 2, 1°, (, i°', tan82— = tan 90°-7 —, 2, I, 2J, <1 r\ *•, , <0, , = cot 7—=(73 +72) (72 +1), 2, , IV. (i) Value of cos 22 —, 2, 1°, Let 0=22—, then 20 = 45°, 2, , _276-273 -4 + 272, 2, , = 76 - 73 - 2 + 72 = 73(^-1), -72(72-1), = (7J-72)(72-1)], , 1° 1 + cos 45°, Now, cos2 22 — =, 2, 2, , 1 + 4=, , __ y[2_, 2, 72 +1 _ 2 + 5/2, 2V2, 4, , www.jeebooks.in, 1

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 1°, V cos 22 —>0, 2, , cos 22 —= -72 + 72, 2, , 2, , 1°, , (, , i°>, , I, , 2J, , 1-, , Again, cos 67---- sin 67 — = V2 sin 45° -67 —, 2, 2, , 1° 1-cos 45°, sin2 22 — =, 2, 2, , 72, , /2 -1, , 2-V2, 4, , 2T2, , 2, , i°, v sin 22 — > 0, 2, , 1° 1 I----- r, sin22-=-72-V2, 2 2, , = -72 sin 22 —<0, 2, 1°, 1°, cos 67---- sin 67 —, 2, 2, , 1°, tan 22 — =, 2, , 2, , 1°, 1°, cos 67---- sin 67 —, 2, 2, , (iii) Value of tan 22 —, 2, 1°, tan 22 — >0, 2, , = -71-Sin 135°, , = i--L, N 72, , 1-cos 45°, , 1 + cos 45°, , 2-1, , 4-2V2, , 2, , 4, , = --74-272, , tan 0 =, , ...(i), , 2, , 1 - cos 20, 1 + cos 20, , Adding Eqs. (i) and (ii), we get, , 2 cos 67 y = | (74 + 272 - J 4-2^2), , i-±, __ V2, , cos67~ = “(74+2V2 -^4-2^2), , 1 + 4=, 72, , =, , (i), , 4, 1°, , (ii) Value of sin 22—, 2, , Similarly, subtracting Eq. (ii) from Eq. (i), we get, , (^-i)2, , = (72-I)2, , \(72 +1)(72-1), , V, , 2-1, , = 7(72 -1)2 =72 -1, 1°, 1°, V. Value of cos 67 — and sin 67 —, 2, 2, , 1°, 1°, r( 1, 1°, 1, i°A, cos 67— +sin67 — = V2 — cos 67 — + —= sin 67 —, 2, 2, 2J, , sin 67 — = — (74 + 272 + ^4 — 2-^2), 2 4, 1°, VI. Value of sin 157 —, 2, 1°, Let 0 = 157— /. 26=315°, 2, Now, sin 0 =, , 1 - cos 20, , 1°, ■: sin 157->0, 2, , 2, , r1, 1°, 1°, = V2 sin 45° cos 67 — + cos 45° sin 67 —, , 2, , I, , 2, , = V2 sinf 45° +67—^ = V2 sin 112 — >0, , <, , 2, , 2>, , Thus, , 1°, 1°, cos 67—+sin 67—, 2, 2, 1°, 1°, cos 67—+sin 67 —, k, 2, 2., , (, , 37, , 1-cos 315°, , 1-cos 45°, , 1- -L, 72, , 2, , 2, , 2, , 1°, sin 157 — =, 2 1, , 4, , 2, , _ 1(72 +1) _ I 4+ 272, 72 V 7J, V2 - 2V2, , 1 + cos 20, 1°, Similarly, cos 157 --2, , www.jeebooks.in, = 1+1, , V, , =--72+>/2, , 1°, V cos 157 — <0, 2

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www.jeebooks.in, 38, , Textbook of Trigonometry, , All these values are tabulated as follows:, , 7.5°, , 15°, , 7T- 2^6 - 2V5, , /3 - 1, , 4, , 2V2, , 7 8 + 276 + 275, , 13 4- 1, , 710+2V5, , 72 + 72, , 4, , 2V2, , 4, , 2, , (73 -72) (72-1), , 2-73, , sin, , cos, , tan, , 18°, , 5-1, 4, , 22.5°, , 36°, , 7T72, , 710-2V5, , 2, , 4, , 5 4- 1, 4, , (73 4-72)(72 4- 1), , 24-73, , 7(5 + 2V5), , 75°, , 2, , 27T, , 72-72, , /3 -1, , 2, , 272, , 72 + 1, , =, , 2+73, , 75-2V5, , 2-1, , 75, cot, , 67.5°, , 2-73, , 2-1, , 1+*, , 75, , I Example 87. show that, , 1 + tan , (n 6 A, _____ 2 = 1 + sin 0 = tan, — 4- - ., COS0, I, 2 2), 1 — tan —, 2, 0, sin -, , 1 + ___ 2_, 0, 0, . 0, cos — cos — + sm —, ____, 2_, _, 2, 2, Sol. L.H.S. =, 0, 0, . 0, sm —, cos---- sin —, 2, 2, , 1, , = 2 + 2 cos( A - B) = 2 [1 4- cos(A - B)], A-B ,, 2 A -B •, = 2x2 cos2 ------- = 4 cos, 2--------------- 2, , I Example 89. Prove that,, 7t, JL, , 371, 44 J/t, , 4 571, , 4 7tt, , 4, , cos4 — + cos----- 1- cos ------- 1- COS, 8, 8, 8, 4 771, , 4 7t, , Sol. L.H.S. = cos — + cos —, 8, 8, 4 71, , cos — + cos, 8, , e, cos —, , it, 71------, , 4, , 8, , 2, , i cos e- 4- sin, . ey, —I, , 2), . 20, cos, - sin -sm, 2, 2, 20, . 2 6 , Q . 00, 0, cos - + sin — + 2 sm - cos —, 2, 2_______ 2, 2 _ 1 + sin 0, 2 ®, -2 0, cos 0, cos---- sin, 2, 2, 0, 7t, 0, 1 + tan tan — + tan —, it 0, 4, Again,------- 1- =, —~~ = tan — + 71, 9, 4 2, 1 - tan - .1 - .tan —, tan —, 2, 4, 2, ,, 0, 1 + tan, 1 +sin 0, Thus ---------- = ---------- = tan —+ —1 = RHS, 0, cos 0, 4 2J, 1 - tan 2, , 2, , 2 A-B, , =2, , =2, , 1 + cos —, 4, 2, , 8, , I +, , cos, 2, , 2, , 4, , 3n, , 7t-------, , 8, , [•/ cos(n - 0) = - cos 0], 2, , 3n, , 8, , r 1 + cos —, 3ti y, , 2, , 4, , +, , 2, , 7, , 1, 2, 1, , 2, , 1 + -U, , 72,, , 1, , 2, , + l--j=, , k, , 72, , 1+i2 +2.4+1+i21 _2.4, , 2, 3, - = R.H.S, 2, , I Example 88. Prove that,, (cos A+cos B)2 4-(sin A+sin B)2 =4cos, , 4 37t, , cos----- 1- cos, 8, , = 2 cos4 — + 2 cos4 —, 8, 8, 2 71, COS —, , 4 57t, , cos — + cos —, 8, 8, , +, , 2, .0, , 2, , 8, 4 371, , +, , 2, , V, , 3, , 2, , 2, , I Example 90. Find the value of tan \, , 2, , Sol. L.H.S. = (cos2 A + cos2 B + 2 cos A cos B), + (sin2 A +sin2 B + 2 sin A sin B), , Sol. Let 0 = —, then 20 = —, 8, 4, , www.jeebooks.in, = (cos2 A + sin2 A) + (cos2 B + sin2 B), , 4- 2(cos A cos B + sin A sin B), , Now,, , tan 20 =, , 2 tan 0, , 1 - tan2 0

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 39, , 2 tan —, • .tan —, n =, 8, 4, 1 - tan2 —, 8, ,, 2x \, 1 =------ r, where x = tan —, 1-x2, 8, , Now, sin2 0 =, , i, , 2^2 - 73 - 1, 4y/2, , r, , sin 0 = sin 7 — > 0, 2, , x2 + 2x - 1 = 0, , =>, , -2 ±2^2, x =-----------2, it, , ■, , v tan — > 0, 8, , y < x < it, then find the, , value of sin -, cos - and tan, 2, 2, 2, , 4, , Sol. Given, a = 112°30', , or, , it < x < n, "422, , = — J8 - 2V6 - 2V2, , I Example 93. If a = 112°3O', find the value Of sin a, and cos a., , X, , XX, , Sol. Here — < x < it, 2, , 2^2 -73-1, 1°, sin 7— =, 4^2, 2, , it, , x = tan- =V2-1, 8, , I Example 91. If tanx = -, , = “(1 “ cos 15°), , ^2^2 - 7? 27i j’2j 2V2, , 1-, , 2, , l-x2=2x, , =>, , =, , 1 - cos 28, 2, , 2a =225°, cos 2a = cos 225° = cos(180° + 45°), 1, = - cos 45° = 2, , x, Hence sin—, cos —, tan — will be all positive., 2, 2, 2, , Now, sin2 a =, , 1 - cos 2a, , 2, Since a lies in the 2nd quadrant :. sin a is positive, , I, sina =, , 4, , 2, + 1, , 4, . it, tan x = — and — < x < it, 3, 2, 3, COS X = —, , 5, , Now,, , . x n, sin —>0, 2, , 1 - COS X, . x, sm — =, 2, 2, , >+-, , 22, __5 _ JL, 2, 75, , 1, x, cos — =, 2, , + COS X, , 2, , 2J, , 1 - cos 2a, , 3, Given,, , ( 1 >, , !, 1, , 1-, , 2, , 2 + 2^2, , V 2^2, , 2, , _^2 4-25/2, , Hence, sin a, , 2, But cos a is negative in 2nd quadrant, h + cos 2a, cos a = -, , 2^5, 5, x n, V cos — > 0, 2, , x, sm —, 5 =—, 1 = V5, x, — and tan — = —2- = 2, 2, 75 5, 2 cos —x, 2, , 23tt, I Example 92. Find the value of sin—., . . 23it . f, n ) .it, n . 1°i°, it, Sol. sm---- =sin| it —- = sm — =sm 7 —, 24, 24, 24, 2, , q, , la—b, , m, , <P prove that,, I Example 94. If tan - = Jtan y,, , 2', , a cos (p+b, cos a -—;----- ., a+b cos (p, 0, Sol. Given, tan — =, 2, , (p, tan—, 2, a+b, a-b, , 1 - tan2 Now, cos 0 =---------- =, 2 8, 1 + tan 2, , a-b, 1-------- tan 2 9, a+b, 2, a-b, 1 +------ tan 2 <P, a+b, 2, , www.jeebooks.in, 1°, Let 0 =7—.then 26 = 15°, 2

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www.jeebooks.in, 40, , Textbook of Trigonometry, , 1 I . „o 1, = -| sin 18° + - 11 - - + cos 36° j, J, 4, 2 \ 2, , sin2—, _2_, a+b, .2 <P, cos —, 2, •, 2, <P, sin —, 1+^, 2, , [•/ cos 72° = cos(90° - 18°) = sin 18°], 1~ 1 ! (75-1)', , ►, , a+b, , _ 1 [(75-1) |, 4, , 2, , 4, , 4, , 4, , (a + b) cos2 — -(a - b) sin2 —, __________ 2____________ 2_, , 1 (75 + 1) (75-1) = (5-1), , —1 + if cos 2 --V sinei«—2, 2, 2, 2J I, , a cos (p + b, , 4, , 4, 4, = --64, , (a + b) cos2 — + (a - b) sin2 —, 2 2, 2 ---<P sm 2 q> + b f cos2 — +sin2 —, a cos, 2, 2, 2, 2, , 64, , 4, , — = RHS, 16, , 7t ., , 271, , 3n, , 4n, , I Example 97. Prove that sin —sin —sin—sin —, 5, 5, 5, 5, 5, ~ 16, , a + b cos q>, , COS a — COS P, , I Example 95. If cos e =, , then prove, , 1-cos a cos p', q oc, B, that one of the values of tan - is tan — cot -., 2, 2, 2, !, , c . 4 z 6 1 - cos 0, Sol. tan - =----------2 1 + cos 0, , 2, , v sin 18° =^-—— and cos 36° __ (75 +1)*, , cos —, , <P +sin, . 2, a cos*2 —, 2, , 2, , 4, , Sol. We have, 4n, TTT„, .7t.2Jt.3jt.4Jt, LHS = sin — sin — sin — sin —, 5, 5, 5, 5, ..........., n , 2n . (, 27T^ . (, n, = sin — sin — sin | ft - — I sin I it, • -—, "" 5 "" 5, V”, • 2^*2 2n, , cos a - cos P, 1 - cos a cos P, cos a - cos P, 1 - cos a cos P, , = sin — sin —, 5, 5, = (sin 36° )2 x (sin 72° )2, , ['.'sin(n - 0) = sin 9], , = (sin 36° )2 X (cos 18° )2, , 1 - cos a cos p - cos a + cos P, , [•/ sin 72° = sin(90° - 18°) = cos 18°], , 1 - cos a cos p + cos a - cos p, , (IO+2V5), , = (10 - 2a/5), , (1 - cos a) + cos P(1 - cos a), , 16, , 16, , (100-20), , “ (16x16), , (1 + cos a) - cos P(1 + cos a), , (1 - cos a)(l + cos p), , J10 - 2V5, sin 36° = - -------4, and cos 18° = ^°^o, , ,a, 2P, tan — cot, 2, 2, , (1 + cos a)(l + cos P), 0, ., a, P, :. tan — = ± tan — cot —, 2, 2, 2, , L, 0, , (x, , 80, , B, , Hence one of the values of tan — is tan — cot —., 2, 2, 2, , I Example 96. Prove that, , (16X16), , 4, , 5, — = RHS, 16, , I Example 98. Find the value of, , cos 6° cos 42° cos 66° cos 78° = —., 16, , Sol. We have, LHS = cos 6° cos 42° cos 66° cos 78°, , (ii) cos22°30', , (i) sin22°3O', (iii) tan 22°30', Sol. (i) sin2 0, , (1 - cos 20), 2, , = — (2 cos 66° cos 6°) (2 cos 78° cos 42°), 4, , sinz(22°30') = j1, = — [cos(66° + 6°) + cos(66° - 6°)], , 4, , 1*, C°S 450 = ;V2., 2, 2, , f(j2-l), , 72-1, 2>/2, , www.jeebooks.in, X [cos(78° + 42°) + cos(78° - 42°)], , = -(cos 72° + cos 60°) (cos 120° + cos 36°), 4, , =>, , sin(22°30') =, , 2>/2

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www.jeebooks.in, Chap 01, , (ii) cos2 0 =, , (1 + cos 20), 2, , cos2(22°30'), , 1 - tan2 2 tan —, 2, ----------- 2- +, , =>, , f, , Trigonometric Functions and Identities, , X^I, 277, , 2, , 2, , k, , 1, 2, , 1 + tan2 2, , 1 + tan2 —, 2, 1 -t2, 2t, x, ,, tan — = t, then, , + l + t,:, 2, 1+t, , (1 + cos 45°), Let, , 7, , 1(77 + 1), N 241, sin2(22°30z) (77 - 1) x (277), (iii) tan2(22°30') =, cos2(22°30')~ (277) (41 + 1), =, = (77-i)x(77-i), 41+ 1 (41 +1) (41-1), = (77-1)2, , =>, , cos(20°30') =, , 3t2-4t-l=0 => t =, , 41, , 1, 2, , 2± 77, 3, , X, , =>, , t = tan — =, 2, , Now,, , 3, , z., , Tt, , It, , X, , v 0 < — < —, tan —, 2, 2, 2, , „, x, 2 tan —, tan x =--------- —, 1 - tan2 2, , => tan(20°30/) =(77 - 1), , f2+4T, , 2, , 2, , <, , 1, I Example 99. If 0 < x < it and cos x+sin x=-, then, , 1-, , tan x = -, , ., , '2 + 41, , \, /,, , find the value of tan x., Sol. Given, cos x + sin x = 2, , 2 + 77, , 3(2 + 77) 1-277, "Xi-277, , [— \, , I 3, , 7, , Exercise on Session 9, ~ 00560 x -s'n x’then, , 1. If tan, , value °ftar|2, , X, , 2, , 2. Find the value of cos41 — j., , 1, 1, 3. Find the value of expression cos 290° + 73 sin 250° ’, , 1 •, , 1, , 4. If x + — = 2 cos 0 then find the value of xn + —., x"', X, , 5. Show that sin 47° + sin 61° - sin 11° - sin 25° =cos 7°., , 6. If a and P be two different roots of equation a cos 0 + b sin 0 = c, then show that sin(a + P) =, , 7. If sin a + sin P = a and cos a + cos p = b, then show that tan, , g-PV ±, 2 ), , 2ab, a2+b2’, , (4-a2 -b2, , a2+b2, , 1°, , 8. Show that cot 142— = 72 + 73 - 2 - 76., 2, , 9. If cos 0 =, , ”, , cos a + cos P then prove that one of values of tan | is tan, 1 + cos a cos P’, , tan, , www.jeebooks.in, 10. Find the value tan - + 2 tan — + 4 cot —., , 5, , 5, , 5

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www.jeebooks.in, 46, , Textbook of Trigonometry, , Exercise for Session 10, , 1. tfA + B + C = 180°, then prove that sin 2A + sin 2 8 + sin2C =4 sin Asin B sin C., , 0, B, C, 2. If A, 4+8, B + C = 180°, then prove that tan2 - = tan — tan —. when cos 0(sinB, 8 (sin 8 + sin C)=sin, C) =sin A., , 3. If A, 8, C are angles of a AABC, then prove that sin 2 A + sin 2 8 - sin 2C = 4 cos A cos 8 sin C., sin A + sin 8 + sin C, k A kB k., _ ..., ., .,, 4. If in a AABC, ------------------------- = 2X cot — cot —, then find the value of A., sin A + sin 8-sin C, 2, 2, cos 8, cos C, cos A, 5. If A + 8 + C = 180 °, then find the value of —*-------- +, + ------------sin 8 sin C sin C sin A sin A sin 8, , 6. In AABC, show that, 7. In a &ABC, if tan, , A, C, 1 - cos A + cos B + cos C, = tan — cot —., 1 - cos C + cos A + cos 8, 2, 2, , B + C-A, C +A-B, A + B-C, tan, tan, = 1, then find the value of cos A + cos 8 + cos C., 4, 4, 4, , 8 + cot —, C = X cot —, A cot —, B cot £, then find the value of X., 8. If, in a AABC,, cot —, “2.2, 1A2 C,A + cot —, 2, 2, 2, 2, 2, 2, 9. If A + 8 + C = —, then show that cos 2 A + cos 28 + cos 2C = 1-4 sin A sin 8 sin C., , 2, , 10. If a + P + y = 2n, then show that tan — + tan - + tan -- = tan — tan - tan -., 2, 2, 2, 2, 2, 2, , Session 11, Maximum and Minimum Values of, Trigonometrical Functions, Conditional Trigonometrical Identities, We have certain trigonometric identities like,, sin20 +cos2 9 = 1 and 1 + tan2 0 =sec20 etc. Such, , identities are identities in the sense that they hold for all, value of the angles which satisfy the given condition, among them and they are called conditional identities., If A, B, C denote the angle of a AABC, then the relation, A + B + C = it enables us to establish many important, identities involving trigonometric ratios of these angles., (i) If A + B + C = 7t, then A + B = it - C,, B + C = ft - A and C + A = it - £, (ii) If A + B + C = 7t, then sin( A + B), = sin(7t - C) =sinC, Similarly, sin(B + C) = sin(7i - A) = sin A, and sin(C + A) = sin(7i - B) = sin B, , (iii) If A + B + C = 71, then cos(A + B), = cos(7t - C) = - cos C, Similarly, cos (B + C) = cos(tc - A) = - cos A, , and cos(C + A) = tan(7t - B) = - tan B, (iv) IfA + B + C = 7t, then tan( A + B), = tan(7t - C) = - tan C, Similarly, tan(B + C) = tan(7t - A) = - tan A, and,, tan(C +A) = tan(7i-B) = -tanB, (v) IfA + B + C = n, then^-£^ = --£, 2, 2 2, B + C it A, , C + A _ rt B, and, --------and, 2, ” 2, 2, 2 2, 2, ' C^, ', a + b\, sin, = sin, = cos, 2,, < 2 J, <2,, ' A + B^, (n C, (cos, = cos, = sin, 2 2, < 2 7, , J, , £, , www.jeebooks.in, I2

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , cos, , (ii), , (iii), i, , i, , (iv), , 2, A+B, , = cos, , 71 _ C ., , 2 2J, , = sin, , -I, , 2J, , u, , = tan — - — = cot, 2, U 2J, All problems on conditional identities are broadly, divided into the following four types :, Identities involving sines and cosines of the, multiple or sub-multiples of the angles involved., Identities involving squares of sines and cosines of, the multiple or sub-multiples of the angles involved., Identities involving tangents and cotangents of the, multiples or sub-multiples of the angles involved., Identities involving cubes and higher powers of, sines and cosines and some mixed identities., , tan, , (i), , A+B, , TYPE I Identities involving sines and cosines of the, multiple or submultiple of the angles involved, I, , Working Method, Step 1 Use C and D formulae., Step 2 Use the given relation (A 4- B + C = 7t) in the, expression obtained in step 1 such that a factor, can be taken common after using multiple angles, formulae in the remaining term., Step 3 Take the common factor outside., Step 4 Again use the given relation (A + B + C = 7t), within the bracket in such a manner so that we, can apply C and D formulae., , Step 5 Find the result according to the given options., TYPE II Identities involving squares of sines and, cosines of multiple or sub-multiples of the, angles involved., , Step 1 Arrange the terms of the identify such that either, sin2 A - sin2 B = sin(A + B) .sin(A - B), or cos 2 A - sin2 B = cos(A + B) cos(A - B) can be, , used., Step 2 Take the common factor outside., Step 3 Use the given relations (A + B + C = 7t) within the, bracket in such a manner so that we can apply C, and D formulae., Step 4 Find the result according to the given options., Type III Identities for tan and cot of the angles, Working Method, , 47, , Step 2 Taking tangent or cotangent of the angles of both, the sides., Step 3 Use sum and difference formulae in the left hand, side., Step 4 Use cross-multiplication in the expression, obtained in the step 3., Step 5 Arrange the terms as per the result required., , ! Example 107. If A + B + C = n, then, find, , sin2A + sin2B + sin2C., Sol. sin2A + sin2B + sin2C, 2A +2B, 2A -2B, = 2sin, cos, + sin2C, 2, 2, , = 2sin(A + B) cos(A - B) + sin2C, = 2sin( n - C) • cos( A - B) + sin 2C, A + B + C = n, A + B = n-C, .'. sin(A + B) = sin(7T - C) = sinC], = 2sinC cos(A - B) + 2sinC cosC, = 2 sinC [cos(A - B) + cosC], = 2sinC [cos(A - B) - cos (A + B)], [•.• cos(A - B) - cos(A + B) = 2 sin A sinB,, by C and D formula!, = 2 sin C [2 sin A sin B], = 4 sin A sin B sin C, , ? Example 108. If A+B + C = it, then, find, tan A + tanB + B + tanC, Sol. A + B + C = n, , A + B = n - C => tan( A + B) = tan(n - C), tan A + tan B, _, ---------------- = - tan C, 1 - tan A tan B, , tan A + tan B = - tan C + tan A • tan B ■ tan C, tan A + tanB + tanC = tanA- tanB- tanC, , Maximum and Minimum Values, of Trigonometrical Functions, As we have discussed in previous article that -1 < sin x < 1, and -1 < cos x < 1., If there is a trigonometrical function of the form, a sin x + b cos x, then by putting a = r cos 0, b = r sin 0, we, have, asinx 4-bcosx = rcos0sinx + rsin0 -cosx, , = r(cos 0 sin x + sin 0 cos x), , www.jeebooks.in, Step 1 Express the sum of the two angles in terms of, third angle by using the given relation, , (A + B + C = n), , = rsin(x + 0), where r = ^a2 +b2, tan0 = a

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities 49, , Application on Quadratic Equations, , As -1 < sin0 < 1, sin0 = 1 - ^1 - X. (neglecting 1 + -Jl - X), , As we know, ax2 + bx + c = 0, represents the quadratic, , . _, 1, From question, sin 0 > - 2, , equation whose,, , Thus,, , --cl-Jl-X <1, , or, , --<-7i-x<o=> Ji - x < -3, , =>, , , , 9, ., 5, 1-X<- => X> ——, 2, 4, , 2, , sum or roots = —., a, a, -b±yjb2 -4ac, (a, P) roots =- -----------2a, and if we want to form quadratic equation whose roots, are given., => x2 - (sum of the roots)x + (product of the roots) = 0., As above mentioned results are basics for quadratic, equations, we discuss certain application on trigonometry., , ct, B, I Example 116. Find cos(a + p), if tan — and tan - are, 2, 2, roots of the equations 8x2 - 26x +15 = 0., Sol. It is given tan — and tan — are roots of 8x2 - 26x + 15 = 0., 2, 2, g, p 13 ,, g, P 15, =>, tan— + tan- = — and tan—.tan- = —, 2 2 4, 2, 2 8, 1 - tan2| g + P Y, 2 J, cos(g + P) =--------g + pY, 1 + tan2l, , I Example 118. If ABCD is a convex quadrilateral such, that 4 sec A + 5 = 0, then find the quadratic equation, whose roots are tan A and cosec A., 5, n, Sol. sec A = —. So, — < A < n, , 4, , or, , tan, , a +P, , 3, , I Example 119. If sec a and cosec a are the roots of, x2 -px + q = 0, then show p2 = q(q+2)., :. sec g + cosec g = p and secg cosec g = q, , =>, , 1+, =>, , sing + cosg = psing cosg and sing cosa. = —, <?, p, sing + cosg = —, <7, , On squaring both sides, we get, , cos(a + P) =, , <676', , oz, , sin2g + cos2g + 2sing cosg =, , 1 + 2sina cos a, , 8, <676', , I 49,, cos(g + P) = —, , 5, , x2 - —x - — = Oor 12x2 - llx - 15 = 0, 12, 4, , 7, , 1-, , ,, , Sol. Since, secg and cosec g are the roots of x2 - px + q = 0, , g, P, tan— + tan2 2, g, P, 1 - tan —tan22, 13, -26, 4, , 2, , 2, , Hence, tan A = — and cosec A = 4, 3, Required quadratic equation is, 3 5], 3 x’=0, x2, —+- X+, 4 3j, 4, 3, , 2 J, , 2, , _________, , 2, , Q, , product of roots =, , g +P, where tan, , __________, , Q, , => p1 = q(q +2), , or, , -627, 725, , I 49,, -627, 725, , I Example 117. If the solutions for 0 from the equation, , sin2 0 -2sin0 + X = O lie in u|2nrt—,(2n+l)n + — ., n ezi, 6, 6,, Then, find the possible set values of X., , ■?2, , 9, , 9, , I Example 120. Find the number of values of x in the, interval [0,5n] satisfying the equation, 3sin2 x-7sinx + 2 = 0., Sol. 3sin2x-7sinx + 2 = 0, 7 ± J49 - 24 7 ± 5, =>, sm x =---- 1---------=------ = 1,2, 3, 6, 6, 1, (where, 2 is not possible)., /., sin x = 3, , www.jeebooks.in, Sol. sin20 - 2sin0 + X = 0 => sin0, , =2±J^=1, , For real values, 1 — X > 0, i.e. X < 1., , 2, , sinx = - = sing;(0 <g < n/2), 3, x = g, n - g, 2n + g, 3k - g, 4n + g, 5n - g, Thus, the number of values of x is 6.

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www.jeebooks.in, 50, , Textbook of Trigonometry, , (x - I)2 + 3 = -3cos(ax + b), , (ii—, , X > - — and X > 1, 4, , Sol. x2 - 2x 4- 4 = -3cos(ax + b), =$, , X>1, X>-1, , =>, or, From Eqs. (i) and (ii),, , 1 Example 121. 0<o< 3, 0<b< 3 and the equation,, x2 + 4 + 3cos(ox + b) = 2x has atleast one solution,, then find the value of (o+b)., , (i), , Eq. (i) is only possible if,, cos(ax + b) = -1 and (x - 1) = 0., a + b = n,3n,5n,..., So,, and, 3n > 6, a 4- b < 6, where, =>, a 4- b = Jt, , X > - — and X > -1, 4, X > 1 or X > -1, X e [-1, °°)., , or, , As -1 < cos(ax + b) < 1 and (x — l)2 > 0, , Proving Trigonometric, Inequality, , I Example 122. Find the values of p if it satisfy;, , Jensen’s Inequality, (i) Suppose that lf is a convex function on [a, &]g R, for, all xp x2, x3..., xn G [a, b],, , cos 0 = x + -, x e R for all real values of 9., x, Sol. x.2 - cos0 x + p = 0, , we have, ^Xj + x2+...+x„', , cos0 ± -Jcos20 - 4p, X=, , n, , 2, cos20 - 4p > 0. => 4p < cos20, , For real x,, , Proof If f(x) is concave up,, , 4p < cos20 < 1., , y., , An(xn. f(xn)), , p < — for all values of0., 4, , G(X.P), , I Example 123. Find the set of values of X g R such, that tan2 0 + sec6 = X holds for some 0., , fo.Jl)), , Sol. tan20 + sec0 = X => sec20 + sec0 - (X + 1) = 0, , *1, , i.e., , 4, or sec0 < -1, Also, sec0 > 1, -1±, + 5 >1, —J4X, y-------2, -1 ±74X4-5, or, <-l., 2, , f(x)), A2(X2, f(X2)), , 0, , -1±71 + 4(X + 1), 2, _ -1 ± yjlX + 5, 2, For real sec0,, 4X + 5>0,, , n, , —(i), , Here,, , G, , and, , P, , 1)), , *X, , X, , 'x, +x2+...+x„ f(xJ+f(x2)+...+f(xn), n, , n, , ' x.i 4-x,+...+x„, l__________ n, , x, + x2+...+xn, , n, , n, , 7, , From figure, ordinate of G > ordinate of P., , /(x1)±f(x2)+...-F/(xn), , n, , J x, + x2+...+xn, , I, , n, , f(x1)+f(x2)+...+f{xn)>n-f, , =>, , or, , 1X + 5 <-2., , =>, or, , 4X + 5>9, 4X4-5 > 1, , (ii) Similarly, suppose that f is concave function on, [r.,b]E R, for all xp x2, x3>... xnn e [a, b], we have, , f(xi) + f(x 2)+.. .+f(xn) < n • f, , x. +x,+...+x„, n, , J, , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , I Example 124. If A, B,C,e, , Then prove that, , =>, , cos A + cos B + cosC, ------------------------- < cos, , A + B+C, , 3, , 3, cos AcosB+cosC <-., 2, , 51, , 3, , cos A + cos B + cos C <, , 3, , as at B + C = it, , 2, y, , Sol. Since, for a function which is concave downwards, /(x1) + /(x2) + /(x3), X, + x2 + X3, f, 3, 3, , L, 1, Q,, , and we know that the graph of y = cos x is concave, , R, , pZ, -n, , downwards for all x 6 I -—, — ., , V 2 2), , 0, 2, , jt, , ■x, , 2, , Let P( A, cos A), Q(B, cos B) and R(C, cos C) be any three, points on y = cos x, then it is clear from the graph, GM < ML, , Exercise for Session 11, 1. Prove that the minimum value of 3 cos x + 4 sin x + 5 is 0., , 2. If sin 0, + sin 02 + sin 03 =3, then find the value of cos 0, + cos 02 + cos 03., 3. If x = r sin 0cos <>, y =r sin 0sin <J> andz =r cos 0, then prove that x2 + y2 + z2 is independent of 0 and 0., , 4. Find the least value of 2 sin2 0 + 3 cos2 0., 5. a, 0, y are real numbers satisfying a + 0 + y = ti. The find the minimum value of given expression, , sin a + sin 0 + sin y., 6. If A = sin2 0 + cos4 0, then find all real values of 0., , 7. Find the minimum value of sec20 + cosec20 - 4., 8. If P = cos (cos x) + sin (cos x), then the least and greatest value of P respectively., , (a) -1 and 1, , 9. Let 0 e [ 0, — I and, 4, , (b) 0 and 2, , (c) -42 and V2, , (d) 0 and 42, , = (tan Qfn e, t2 = (tan 0)' 9, t3 = (cot 0/®"9 and f4 = (cot 0)“*8, then show that f4 > t3 > t, > f2., , 10. Find the ratio of greatest value of 2 - cos x + sin2 x to its least value., , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , • Ex. 6. The sum ——— --------- +-------------------sin 45° sin 46° sin 47° sin 48°, , So/. (a)/(0) =, , Sol. (b) T\ =, , _ (cosQ -sin0)2 + (cos20 -sin29), , =, =, ■, , T, 12 —, , 2(cos0 -sin0)(cos0 +sin0), _ (cos0 - sin0) + (cos0 + sin0), , (b)cosec (1°), (d) None of these, , sin(46° - 45°), 1, sin 1° sin 45° sin 46°, , 1, , sinl0, , sinl0, , 2(cos0 + sin0), , /(ll°)-/(34°) =, , 1, , 1, 1, (1 + tanll0) (1 + tan34°), 1_______, , 1, , [cot 47° - cot 48° ], , 1, (1 + tanll0) ! ! 1- tanll0, 1 + tanll0, 1, , sin 1° sin 133° sin 134°, , sin 1°, , 1, 1 + tan0, , (1 +tanll0) (1 + tan(45° - 11°)), , sin(133° -134°), , 1, , _, , 2cos0, , sin(48°-47°), sin 1° sin 48° sin 47°, 1, , 2(cos0 + sin0), , [cot 45° - cot 46° ], , 1, , ■, , 1 - sin20 + cos20, , 2 cos20, , 1, 1, 4-------------------- +... d---------------------- is equal to, sin 49° sin 50°, sin 133° sin 134°, , (a) sec (1°), (c) cot(1°), , 53, , 1, , [cot 133° - cot 134°], , (1 + tanll0), , 1 4- tanll0 _ 1, 2, ~2, , On adding, I, sin 1°, , • Ex. 9. The variable ‘x’ satisfying the equation, , [{cot 45° + cot 47°, , |sin xcos x| + -j2 + tan2 x+cot22 x = Ti, belongs to the, , + cot 49° + ...+ cot 133°}, - {cot46° + cot48° + cot50° + ...+ cot 134°}], = cosec 1°, [all terms cancelled except cot 45° remains], , Ex. 7. The range of k for which the inequality, , interval, , (a) 0,-, , <*” 1.5, , (c) —, , (d) Non-existent, , L4, , k cos2 x - k cos x +1 > 0 V x e (-<», °°), is, , Sol. (d) [sin x cos x| + |tan x + cot x| =, , (a)*<v, , (b) k < 4, |sinx cosx| +, , (c) — < k < 4, 2, , (d)-<k<5, 2, , k cos2x-fccosx + l>0V xe(, =>, , cos2x - cosx =, , 1', , 2, , 4, , f(x) = cos x + cos(x + a) + cos(x + 2a) takes some constant, number c for any xE R, then the value of [c + a] is equal to, , =>, , - — < cos2x - cosx < 2, 4, k, :. We have, 2k +1 > 0 and — +1 > 0, 4, , Hence,, , • EX. 8. If /(0) =, , Note [y] denotes greatest integer less than or equal to y., , (a)0, , (c)-1, , (b)1, , (d)2, , Sol. (d) f(x) = cosx + cos(x + 2a) + cos(x + a), , - - < k < 4., 2, 1 - sin 20 -l-cos 20, , 1, [sin x cos x|, , • Ex. 10. Let a be a real number such thatO < a < 7t. If, , COS X-----, , 2, , |sinx cosx|, , Hence, no solution., , k(cos2x - cosx) + 1 >0, , But, , |sinx cosx|, , but, , Sol. (c) We have, , 1, , = 2cos(x + a)cosa +cos(x + a), = (2cosa + l)cos(x + a), , then value of, , 2cos20, , As cos(x + a) can take any real value from - 1 to 1, Vx 6 R, f(x) is constant, so (2cosa + 1) = 0 must hold., 2ft, , ,, , www.jeebooks.in, /01°) • /(34°) equals, , 1, (a)“, , 3, (b)^, , 2, , 4, , a = — and c = 0, 3, , (c)7, , (d)1, , Hence,, , r, , •>, , 2ft, , [c + a]= 0 + y = 2

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www.jeebooks.in, 55, , Chap 01 Trigonometric Functions and Identities, , I, , (a) sin a, sin b, sin c, , • (271 ], , sml —, 115 ), , (b) cos a, cos b, cos c, , (c) sin 2a, sin 2b, sin 2c, , 16, , 16-sin —, , 115., , (d) cos 2a, cos 2b, cos 2c, , Sol. (b) Equation first can be written as, x sin a + y x 2 sin a cos a + z x sin a(3 - 4 sin2 a), , Therefore, tan-1 P < tan-1 S., • Ex. 16. Set of values of x lying in [0, 2ti] satisfying the, , inequality \ sin x | > 2 sin2 x contains, , . X(A n), , (, , 77t'l, 77t, , \ 6J, , \, , 6 ), , (a), , i, , (c)7, , = 4 cos a(2 cos2 a - 1) as sin a * 0, , (b)l°’T, , =>, , (d) None of these, , =>, , 7 U pt, —, , . '71, , =*, , Sol. (a) | sin x | > 2 sin2 x, , => 0 < | sin x | < I, , n1, , tc, f 55rt, , => X G 111., 0, — kJ ---- , Tt, , 6, , 'I1, , 8 cos3 a - 4z cos2 a -(2y + 4) cos a + (z - x) = 0, 3, , I Z, , y+ 2, , U, , 4, , cos 3 a - - cos2 a cos, , z - X, , cos a +, , 8, , =0, , which shows that cos a is a root of the equation, z'l, y+z, z-x, t’ = - k 2, =0, k+, 4, 8, .2;, , => |sin x | (21 sin x | - 1) < 0, , ■, , = 2 x 2 sin a cos a cos 2a, x + 2y cos a + z(3 + 4 cos2 a - 4), , (, , 7n\, 771, , f 1171, 117t „, , U Tt, ---KJ ------, 27t, ,11., , 6, , Similarly, from second and third equation we can verily, that cos b and cos care the roots of the given equation., , 6, , • Ex. 20. Leto, andfi be any two positive values of x for, , • Ex. 17. The number of ordered pairs (x,y), when, x, y e [0,10] satisfying, , I--------------------------- A, . 2, •, ,1, a, sin x-smx + - -2 “S'<1 is, , minium value of\ a + 01 is, , 2J, , (a)0, , (b)16, , (c) infinite, , (d) 12, , 1 |2 1 1 w, sin x — + - > -, vx, 2 2, 2 1, , •- I 1, Sol. (b) sin2 x - sin x + - =, , 2 ], , '' V, , which 2 cos x, | cos x | and} - 3 cos2 x are in GP. The, , and sec2 y > 1, Vy, so 21" 7 > 2. Hence, the above inequality, 1, holds only for those values of x and y for which sin x = 2, and, sec2 y = 1., 7t 571 137t 1771, ,, n n „, Hence, x =—, —,---- ,----- and y = 0,7t, 27t, 3tc. Hence,, 6 6 6, 6, required number of ordered pairs are 16., • Ex. 18. The least values ofcosec2 x + 25 sec2 x is, , (a) 0, , (b) 26, , (c) 28, , (d) 36, , (b)4, (d) None of these, , Sol. (d)2 cos x, | cos x |, 1 - 3 cos2 x are in GP., cos2 x = 2 cos x ■ (1 — 3 cos2 x), , =>, , 6 cos33 x + cos2 x - 2 cos x = 0, , 1, 2, cos x = 0, -, —, 2 3, 71 7t, ( 2, x=, cos, 2 3 <, 3, If, , 71 Q, , 71, , a = -,p = 2 3, , ft, , Then, | a - £ | = —, 6, n, , • Ex. 21. Letn be an odd integer. If sin nQ = ^br sin' 0, , Sol. (d) cosec2 x + 25 sec2 x = 26 + cot2 x + 25 tan2 x, = 26 + 10 +(cot x - 5 tan x)2 > 36, • Ex. 19. Ifx sin a + y sin 2a + z sin 3a = sin 4a, x sin b + y sin 2b + z sin 3b = sin 4b, , [va.P are positive], , r=0, , for all real 0, then, , (b) b0 =0, b, = n, (d) b0 = 0, b, = n2 - 3n - 3, , (a) b0 = 1, b, = 3, (c) b0 = - 1, b, = n, n, , Sol. (b) Given, sin n0 = ^br sin' 0 = b0 + b;sin 0 + b2 sin2 0, r-0, , www.jeebooks.in, Then, the roots of the equation, , t3, , 2, , A, , 2, , y+2, , I 4, , t+, , z-X, , 8, , +... + b„ sin" 0 ...(i), , = 0,a,b,c* rm, are, , Putting 0 = 0 in Eq. (i), we get 0 = bQ

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www.jeebooks.in, 56, , Textbook of Trigonometry, , n, , Again, Eq. (i) can be written as sin n0 = ^fbr sin' 0, r-0, , sin0, , lim n, , V, , ----- ,--------- , ------, , 4, n, \, 0 6(0, 71), 4, , = ±b, sin'" 0, Hence,, , J ^sin 0, , (, , 71, 7t^ ., ----- <x<------ , then cos 2x, , 4J, , I 2, , is equal to, , (a) a2, , n = bt, bo=O-,b} = n, , (d)n7(2-n!), , (b) a7(2 + a), , (b) {b+c,b-c}, (d) None of these, , jvIt, , TC, , 7T, , 9 Ex. 22. The minimum and maximum values of, ab sin x + by/(l - a2) cos x + c (| a | < 1, b > 0) respectively are, (a) {b - c, b + c), (c) {c - b, b + c}, , 2, I 7t 71, 4 2, , • Ex. 24. Ifcos x + sin x = a, , sin 0 j f 0 ', , e-»o, , Hence,, , 7t 57t 71, , But, , On taking limit as 0 —> 0, we get, sin n0 = bi, hm-------sin 0, , =>, , LHS = RHS, , Hence,, , Sol. (d)‘: — < x <---- 1 - it < 2x< - —, i.e, in III quadrant, 2, 4^, 22, , =>, cosx+sinx = a, Squaring both sides cos2 x + sin2 x + 2 cos x sin x = a2, sin 2x = (a2 - 1), , =>, , Sol. (c) absin x + byj(l - a2) cos x, , cos 2x =71 ~(a2 - I)2, , Now, 7(^)2+(&7(i^7)y, , =7a2(2-a2), , = y]a2b2 + 62(1 - a2), , = a7(2 - a2), , = by](a2 + 1 - a2) = b, , 2 2n, 2 n, • Ex. 25. IfS = cos —+cos —- + ... + COS 1, 2, n, S equals, n, 1, (a), (b), 2(n +1), 2(n -1), , b{(a sin x + 7(1 - a2) cos x)}, , =>, , Let,, , a = cos a,, 7(1 - a2) =sin a, , =>, , 6 sin (x + a), , -1 <sin(x+ a) < 1, c-6<hsin(x + a) + c<6 + c, 6sin (x + a) + c e [c - b, c + 6], , (c), , sin0 -cos0, , 7(1 +cot2 0), , 2, , -2 tan0cot0 = -l, 0G [0,27t], then, I It, , (b)0e -,n, I2, , (c)esH)’M, Sol. (d), , sin’ 0 - cos’ 0, sin 0 - cos 0, , 4 4, = 1 + sin 0 cos 0, cos 0, and, 7(1 + cot2 0), , (d)0e(O,7t)-, , (<, 2, , 2(n - 2), , 2tt, Sol. (c) S = cos2 — + cos2 — + + cos 2/(n - «1)\ —, n, n, n, ,, 2rt ,, 4it, 6n, 1 + cos — +1 +cos — +1 + cos —, 1, n, n, n, , COS0, , • Ex 23., , 1, , then, n, , Tt, , +, , + 1 + cos 2(n - 1) —, n, , 3n, , 1, 2, , 4, 4 2, , n-1, , 2kit, , "-i + S cos---n, k-1, , = l(n-l-l)=l(n-2), , = sin2 0 + cos2 0 + sin0 cos 0,, , • Ex. 26. Ifcos 50 = a cos 0 + 6 cos3 0 + c cos5 0 + d, then, (a) a = 20, (c) a + 6 + c = 2, , (b) b = - 30, (d)a + 6 + c + d = 1, , It, , cos 0, , | cosec 0 |, , = sin 0 cos 0 V 0 6 (0,7t), , Sol. (d) Put 0 = — in the given inequality, we get d = 0, 2, Put 0 = 0 in the given inequality, we get, a+b+c+d=l, So, (d) is correct and (c) is not correct., , www.jeebooks.in, 71, , and - 2 tan 0 cot 0 = - 2,0 * —, 2, , .(0

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , Now differentiate both sides with respect to 0, we get, - 5 sin 0 = - a sin 0 - 3 b cos2 0 sin 0, , Sol. (b) AB = 0, , - 5c cos4 0 sin 0 ...(ii), , cos2 a, , cos a sina, , COS2 P, , cos P sin p, , cos a sin a, , sin2 a, , [cos P sin P, , sin2 P, , 0 0, ~ 0 0, , 71, , ••■(Hi), 0 = —, then a = 5, 2, 7t, Again putting 0 = — in the given expression or in (2), we, 4, get, 4a + 2b + c = - 4, -(iv), From (i), (iii) and (iv) we have b= - 20 and c = 16, [Note We have found correct answer at the second step, only however the complete solution is desired for better, understanding of the solution.], Alternates Solution, , Put, , cos 59 = cos(26 + 30) = cos 29 cos 30 - sin 20 sin 30, =(2 cos2 0 - 1) (4 cos3 0 - 3 cos 0), - (2sin 0 cos 0) (3 sin 0 - 4 sin3 0), , = 8 cos5 0-10 cos3 0 + 3 cos 0 - 2(1 - cos2 0), , 57, , cos a cos P cos(a - P), , cos a sin P cos(a - P), , cos P sin a cos(a - p), , sin a sin P cos(a - p), , 0‘, , 'o, , ” .° °., => cos(a - P) =0, , • Ex. 29. Ifk} = tan 270 - tan 0, ,,, sin 9, sin 30, sin 90 ,, and k2 =--------- +---------- +------------then,, cos 30 cos 90 cos 270, (a) ky=k2, , (b) k, =2k2, (c) k, + k2 = 2, , cos 0{3 - 4(1 - cos2 0)}, , = 8 cos5 0-10 cos3 0 + 3 cos 0 - 2(cos 0 - cos3 0), , (4 cos2 0 - 1) = 16 cos5 0 - 20 cos3 0 + 5 cos 0., , (d) k2 = 2k,, , Sol. (b) We can write, k, = tan 270 - tan 90 + tan 99 - tan 30 + tan 30 - tan 0, But tan 30 - tan 0 =-------------------------------cos 30 cos 0, , equations 3 sin 2 A + 2 sin2 B =1 and, , sin 20, , 3 sin 2A - 2 sin 38 = 0, then A + 2B is equal to, , (a) 5, , (b)S, , . . 3n, , ... 2it, , sin 30 cos 0 - cos 30 sin 0, , n, , _, • Ex. 27. If A and Bare acute positive angles satisfying the, , cos 30 cos 0, 2 sin 0, , cos 30, , (c)T, , (d)-, , fc,=2, , Sol. (b) From the given relations, we have, ( 3^, sin 2B = - sin2A and 3 sin2 A = 1 - 2 sin2 B = cos 2B, , 12 J, , so that, cos(A + 2B) = cos A cos 2B - sin Asin 2B, 2, f 3^, = cos A • 3 sin A - - sin A sin 2A, , sin 90, , cos 270, , sin 30, , sin 0, , cos 90, , cos 30, , +------ +, , • Ex. 30. Ifa2 - 2a cos x +1 = 674 and tan, , 'x, <2, , = 2k2, , = 7 then, , the integral value ofa is, , (b) 49, (d) 74, , (a) 25, (c) 67, , x'l, , = 3 cos A sin2 A - 3 sin2 A cos A = 0, , A +2B = —, 2, , • Ex. 28. If A =, , B=, , 1 - tan2, 2_)_, Sol. (a) 674 = a2 - 2a----+1, x^, 1 + tan2, , 2J, , cos2 a, , cos a sina, , cos a sina, , sin2 a, , COS2 P, , cos P sin P, , cos P sin P, , sin2 3, , and, , =>, , 2 n, 1 - 49 ,, = a2 - 2a x--- + 1, 1 + 49, 2 n, 48 ,, = a + 2a x — + 1, 50, 25a2 + 48a - 673 x 25 = 0, , www.jeebooks.in, (a- 25) (25a + 673) = 0, , are two matrices such that AB is the null matrix, then, , (a) a = P, (c) sin(a - p) = 0, , (b) cos(a - P) = 0, (d) None of these, , =>, , a = 25, , (taking the integral value of a).

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www.jeebooks.in, 58, , Textbook of Trigonometry, , • Ex. 31. The maximum value of(cos a,) (cos a2)..., , or, , 71, (cosan) under the restrictionO <avCL2,...,an <—and, , A =v2 cos — sin x + sin — cos x, 4, 4, 71, , = f?. sin - + X, 4, , (cota,)(cota2)...(cota„) =1 is, , Again, by Eq. (i), 71, , 71, , A =v2 sin — sin x + cos — cos x, 4, 4, , 22, (d)1, , (c)7", 2n, , /71, =V2 cos----- x, 4, , Sol. (a) From the given relations we have, n, , n, , Fl (cos a() = FI (sin aj, n, , 7T, , n, , n, , 11 (cos2 a.) = n (cos a. sin a,)= Fl, , /=]!, , 1=1, , /■I, , • Ex. 33. LetO < 0 < — and x = X cos 0 + Y sin 0,, 2, y = Xsin0-K cos® such thatx2 +2xy +y2 -aX2 +bY2,, , sin2a(, 2, , where a and b are constant, then, , Since,, , =>, , 0<a < — => 0 <2a, < n, 2, ", 1, FI (cos2 a() < — as max. value of sin 2a ( is 1 for all i., , • Ex. 32. The value of expression, , sin3 x, , sin 20 - 2Xy cos 20, = (1 + 2 sin 20)X2 +(1 -2sin20)yz-2cos20 - Xy, , 3, , cos x, -I-------------1 + cos x 1 - sin x, , From the question,, a =1 + 2sin20, b = 1 - 2 sin 20, cos 20 = 0, , is/are, , Tl, , n, (a) Vi cos -----x, , cos 20 = 0 => 0 = —, then, 4, , 7t, (b) V2 cos +X, 4, , 4, 71, , 4, , sin3 x, , 7T, , 71, , a = l+ 2 sin—, b = l- 2sin —, 2, 2, a = 3, b = - 1, , (d) None of these, , ------ X, , cos3J x, , Sol. (a) Let:--------- +--------- = A, then, 1 + cos x 1 - sin x, , 7t, , • Ex. 34. lfO<x < — andsinn x + cos'* x > 1, then, , 2, , (sin3 x + cos3 x) + (cos4 x - sin4 x), (1 + cos x)(l -sinx), , zi —-----------------------------------------------------------------------------, , (a) n e [2, “a) (b) n e (7E, , (cos x +sin x) (cos x - sin x), , +, (cos2 x +sin2 x), A=(1 + cos x)(1 - sinx), (sin x + cos x) {(1 - sin x cos x), + (cos x - sin x)}, A=, 1 + cos x - sin x - sin x cos x, , A = sin x + cos x, 1 ., 1, sin x +, cos x, V2, V2, , 2], , (c) n e [- 11] (d) None of these, , {(sin3 x + cos3 x)}, , or, , (d)9 = 5, , x2 + 4xy + y2 =X2 + y2 + 2(X2 - y2), 22, , or, , (c) a = 3, b = - 1, , Sol. (c) x2 +y2 = X2 + F2,, xy =(X2 - y2)sin 0 • cos 0 - XY(cos2 0 - sin2 0), , So, the maximum value of the given expression is, , or, , (6)9=5, , n(cosa()<422, , or, , (a)a = -tb = -3, , -.(i), , Sol. (b) Since, 0 < x < —, 2, 0 < sin x < 1 and 0 < cos x < 1, when x = 2, sin” x + cos" x = 1, when n> 2, both sin" x and cos" x will decreases and hence, sin" x + cos" x < 1., when n > 2, both sin" x and cos" x will increase and hence, sin" x + cos" x > 1., Thus, sin" x + cos" x > 1 for n < 2., , www.jeebooks.in

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www.jeebooks.in, 59, , Chap 01 Trigonometric Functions and Identities, 3 1, 1 + cos 40, 7, 1, .A, =-+= — + - cos 40, 4 4, 2, 8 8, , • Ex. 35. Ifa = sin — sin — sin —, and x is the solution, 18, 18, 18, the equation y = 2[x] + 2 andy =3[x - 2], where[x], denotes the integral part ofx, then a is equal to, (a)[x], , (b)-L, , (c)2[x], , (d)[x]2, , [*], , 3, -<A<1 => f\-\<A<f(Q), 4, 14;, , • Ex. 37. The value ofcos — + cos — +cos — is equal to, (a) 1, , (b)-1, , Sol. (b) a ~ sin — sin — sin —, 18, 18, 18, = sin 10° sin 50°sin 70°, , (0-2, , = i [2 sin 70° sin 10° ] sin 50°, , _ . ,., (2n I, f 4n, ( 671'l, Sol. (a) cos — + cos, + cos, I 7 J, 7 ,, I, , 7, , 2xi, , =, , [cos 60° - cos 80° ] sin 50°, , = — sin 50° - — (2 cos 80° sin 50°), 4, 4, , (b)/(0)<A</(-2), , (c)/^<x</(0), , (d)/(- 1)<A</(-2), , Sol. (a) Given, /(x) = ax + b, f'(x) = a, Since, a < 0, /(x) is a decreasing function, f(- 1) = 2 and /(I) = 0, - a + b = 2 and a + b = 0, a = - 1 and b = 1, Thus,, /(x) = - x + 1, fl'I 3, Clearly,, =, 7 =7J(-2) = 3>, k 4) 4, , 6*1, , - 1 + 1 + e 7 + e~, ____ s______, , + eT, , 2, , -2*1, , 7, , -4*1, , -6*1, , +e 7 + e 7, , 2, , -1+0, , 1, , 2, , 2, , • Ex. 38. The number of integral value of k for which the, equation 7 cos x+5sinx = 2fc+1 has a solution is, , (b) 8, (d) 12, , (a)4, (c) 10, , Sol. (b) Since, - ^a2 + b2 < a sin x + b cos x < -Ja2 + b2, , .’., , - ^/74 < 7 cos x + 5 sin x < 774, , So,, , - ^74 < 2k + 1 < 774, , 2k + 1 = + 8, ± 7, + 6,± 1, 0, k = - 4, ± 3, ± 2, ± 1,0, so, 8 values of k., , Therefore,, So,, , • Ex. 39. Ify =, , sin4 x-cos4 x + sin2 xcos2 x, sin4 x+cos4 x + sin2 xcos2 x, , x G 0, — , then, , I 2j, ,, , V, , 3, , ., , ., , J, , (b)1<y<2, , 2, (c) - | < y < 1, , /I- =p('1)=2, , 2*1, , 4X1, , [x], , (a)/[2]<x</(o), , - 6*1, , -4*1, , 2, - 1 + (Sum of seven roots of unity), , 1, a=—, , is such that, , -4*1, , 6*1, , 2, ', , y = 2 [x] + 2 and y = 3 [x - 2], => 2 [x] + 2 = 3 [x - 2], = 3 [x] + 3 [—2] => [x] = 8, , [-1,1] o/?to[0,2], then for all values ofQ, A =cos2 0 + sin4 0, , 4*1, , 2X1, , e~ + e 7 + e 7 + e 7 + e 7 + e 7, , = — sin 50°- —(sin 130° - sin 30°), 4, 4, 1, = — sin50° - — sin 50° + — • 4, 4, 4 2 8, , • Ex. 36. If the mapping f(x) =ax + b,a<0 and maps, , 6X1, , 4*1, , = Re e~ + e 7 + e 7, , (d) None of these, , 3, , 1 + cos 20, Also, A = ------------ +, 2, , 1 - cos 20, , 2, , Sol. (d) y =, , sin4 x - cos4x + sin2 x cos2 x, , www.jeebooks.in, 2, , 1, = - + - cos 20 + — - - cos 20 + — cos2 20, 2 2, 4 2, 4, , sin4 x + cos4 x + sin2 x cos22 x, , (sin2 x - cos2 x) (sin2 x + cos2 x) +sin2 x cos2 x, , (sin2 x + cos2 x)2 - sin2 x cos2 x

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www.jeebooks.in, 62, , Textbook of Trigonometry, , • Ex. 46. If fl cos A = cos B + cos3 B, and, , x,, x2 c-a, tan —- • tan — =-----2, 2, c+a, , yfl sin A = sin B - sin3 B then sin(/4 - B) =, (a)±i, , (b)±^, , (e)±i, , (d)±], , 1 - tan — tan —, 2, 2, 2b, , 4, , Sol. (c) V2 cos A = cos B + cos’ B, , and, , tan — + tan —, 2, 2, , Thus, tan, , ...(i), , flsinA = sinB-sin’ B, , -(ii), , = - sin B cos B, , c+a, , • Ex. 48. The minimum value of the function, sin x, , ,, v, , =} sin(A-B) = —Lsin2B, 2V2, , f(*) =, , Now squaring and adding Eqs. (i) and (ii), we get, 2 = cos2 B + sin2 B + cos6 B + sin6 B, , +, , + 2(cos* B -sin4 B), , 1 = (cos2 A + sin2 A)3 - 3 cos2 A sin2, , A(cos2 A + sin2 A) + 2 cos 2B, , cos x, , 71 - cos2, tan x, , 7sec2, , ■yi - sec2, , x, , cot x, , ---- +, , x, , whenever it is defined is, , cosec2 x-1, , x-1, , (a) 4, , (b)-2, , (c)0, , (d)2, , Sol. (b)f(x) =, , ( 3^, , sin x, y/1 ~ COS2, , COS X, , X, , 1 = 1- — sin2 2B + 2 cos 2B, , yfl-■ sin2 x, , W, , +, , - 3 sin2 2B + 8 cos 2B = 0, , =>, , 3 cos2 2B + 8 cos 2B - 3 = 0, 1, cos 2B = -, , =>, , =, , cot X, , 2 '~, , 7, , y cosec x-1, , - 2, x e 2nd quadrant, , 0, x e 3rd quadrant, - 2, x G 4th quadrant, , sin( A - B) = ± 3, , /Wmin. = ~ 2, , • Ex. 47. If xy and x2 are two distinct roots of the equation, *1 +*2, , 71, , • Ex. 49. IfO <a< —, thena(coseca) is, , is equal to, , 2, , 7C, , (b)6, a, , 7C, , (a) less than —, 6, , (b) greater than —, 6, , 7C, (c) less than y, , (d) greater than —, 3, , (d)-, , a, , ■Jsec2 x-1, , + /, , 4, x e 1st quadrant, , 2V2, sin 2B = ± —, 3, , (C)£, , tan x, , sin x, cos x, tan x, cot x, =---------- +----------—F ------------ F, | sin x |, | cos x |, | tan x | I cot X I, , 3, , a cos x + b sin x = c, then tan, , a, , c-a, , 1-, , fl, sin A cos B - fl cos A sin B, , = (sin B - sin’ B) cos B -(cos B + cos’ B) sin B, , bf, , c+a, , c, , Sol. (c) In the graph of y = sin x. Let, , Sol. (b) acos x + b sin x = c, , A s (a, sin a), B = | —, sin —, a 11 - tan2 —, , V£, 1 + tan2 2, X, , 2b tan —, +---------- — = c, X, 1 + tan2, 2, , 16, , 6, , y, , X, , (c + a) tan2-----2b tan — + c - a = 0, 2, 2, x., x., 2b, tan — + tan — =------ ,, 2, 2, c+a, , www.jeebooks.in, 0, , a, , n/6, , ■x

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www.jeebooks.in, 63, , Chap 01 Trigonometric Functions and Identities, , Clearly, slope of OA > slope of OB, so, . 7t, o„, 71, sin a > sm 6 _ 33, a <—, ., 3, 7t, a, 7t% sin a, 6, , Sol. (d) In the second quadrant, sin x < cos x is false, as sin x, is positive and cos x is negative., In the fourth quadrant, cos x < tan x is false, as cos x is, positive and tan x is negative., In the third quadrant, i.e. —, — if tan x < cotx then, 14 2 J, tan2 x <1, which is false., , • Ex. 50. In which one of the following intervals the, inequality sin x < cos x < tan x < cot x can hold good?, , v ( 7tt „, (a) —, 271, , I, , ,.x I 37t, (b) —, , I4, , ,. (5n 3ti, , I 4J, , *, , k 4, , also true., , (d) p.7, I 4, , (c) T’T, , 71 1, , Now, sin x < cos x is true in 0, — and tan x < cot x is, , Further, cos x < tan x, as tan x = -——- and cos x < 1., (cos x), , JEE Type Solved Examples:, More than One Correct Option Type Questions, 1, • Ex. 51. Ifx e (0, tc) and cos x + sin x = -, then tan x is, 2, equal to, , (a), , (c), , 4-77, , , ., , -(4 4-77), , (d), , 3, , 3, , 1, Sol. (c,d) Given, cosx -t-sinx = 2, 1, =>, 14-sin2x = —, 4, 3, , =>, , 4, =>, , 2x e (ti, 2tt), , (, , tanx <0, , X6 — ,71, , 2t, , 3, , 14-t2, , 4, , ---------- zz —, , 7, , 7t, , 7z, , 14, , 14, , 7t, 271, . 271, 1 4- COS — 4- COS----sin—, (c)------- ~~, <d>—, V—, . 71, . 27t, 1 - cos —, sin— 4- sin —, 7, 7, 7, 7t, It, 2tt, 4n, 871, Sol. (a,c,d) tan — +2tan— + 4 tan— + 8cot— = cot—, ', ' ", 7, 7, 7, 7, 7, _ f, , /, , 7t, , [tan0 4-2tan20 4- 4 tan 40 4-8cot80 = cot0 when0 = —], 7, 1 + cot20, n, n, (a) cosec 20 + cot 20 =, = cot0 = cot—, sin 20, 7, , 71, , _., , 2-3, , -8±728, 2-3, , -44-77, , 7C, , „, , 7t, , 71, , „, , (b) tan----- cot— = - 2 cot—, 14, 14, 7, . 2tt, „ . 71, 7t, sm —, 2sm — cos —, 7t, 7, 7 _ cot —, (c) ------- 7, 2n, 7, 2 sin2—, 1 - cos —, 7, 7, 271 i, , (, , -(44-77), 3, , or, , 7, , z,, zt x, , (b) tan----- cot—, (/, , 8f = - 3 - 3f2, , _ - 8 ± 764 - 36, , t=, , 271, , (where, 0 = —), =>, , => 3t2 4- 8t 4- 3 = 0, where t = tan x, , t=, , 271, , (a) cosec—4- cot—, , (b), , 3, , sin2x = -, , • Ex. 52. The value of the expression, 7C, 2.TC, 47t, 8tc, tan — + 2 tan — 4-4 tan — 4- 8cot —- is equal to, 7, 7, 7, 7, , (d), , I, , 1 4- cos— 4- cos —, , n •, , 2, , 7J, 7t, , ., , 71, , 2sm — cos — 4- sm —, 7, 7, 7, , 2^, , n, , 2cos — 4- cos —, 7, 7, . ., , 711, 71, , 71, , 2sm—I cos — 4-1, 7, 7, , www.jeebooks.in, 3, , 71, , = cot—, 7

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , • Ex. 57. Which of the following statements are always, , =>3(1 - m2) tan2 0 +(4m2 + 6m - 4) tan 0 - 8m = 0, , correct? (where, Q denotes the set of rationals), , =>, , (a) cos 20 G Q and sin 20 G Q => tan0 G Q (if defined), (b) tan0 G Q => sin 20, cos 20 and tan 20 G Q (if defined), , 65, , (3 tan 0-4) [(1 - m)2 tan 0 + 2m] = 0, , which is true if tan 0 = - or tan 0 = ———., 3, (m2 - 1), , (c) If sin0 G Q and cos0 e Q => tan30 G Q (if defined), , (d) If sin 0 e Q => cos30 G Q, , • Ex. 60. Ifx cos a + y sin a = x cos p, (, 71A, + ysinP = 2a 0<a,p<— ,, , Sol. (a,b,c), , (a) tan0 =, , 1 - cos 20, , <, , => (a) is correct, , sin 20, , (b) sin 20 =, , tan 20 =, (c) tan30 =, , ., 1-tan20, ; cos 20 =-------- —;, 1 + tan20, 1 + tan 0, , ■2tan0, , 2tan0, 1 - tan20, sin30, , 2J, , (a) cos a +cos P =, , then, , 4 ax, , x2 + y2, , /L\, o 4fl2 - y2, (b) cos a cos P = —-——, x2 + y2, , => (b) is correct, , (c) sin a + sin 0 =, , => (c) is correct, , cos 30, , (d) sin0 = - which is rational but, 3, cos30 = cos0(4 cos20 - 3) which is irrational => (d) is, , incorrect., • Ex. 58. In AABC, tan B + tan C = 5 and tan A tan C = 3,, , (d) sin a sin p =, , x2 + y2, , 4a2 - x2, , x2 + y2, , Sol. (a, b, c, d) We find out the given relations that a and p, are the roots of the equation, x cos0 +y sin0 =2a, (x cos 0 - 2a)2 =(- y sin 0)2, , =>, , then, , 4ay, , =>, , x2 cos2 0 - 4ax cos 0 + 4a2, , (a) AABC is an acute angled triangle, , = y2 sin2 0 = y2(l - cos2 0), , (b) AABC is an obtuse angled triangle, , (x2 + y2)cos2 0 - 4ax cos 0 + 4a2 - y2 = 0, , (c) sum of all possible values of tan A is 10, , =>, , (d) sum of all possible values of tan A is 9, , which, being quadratic in cos 0, has two roots cos a and, cos p, such that, 4 ax, cos a + cos P =, x2 +y2, , Sol. (a,c) tanA + tanB+ tanC = tan A tanB tanC, =>, tan A + 5 = 3tanB, =>, 5 + tanA = 3(5- tanC), 9, 5 + tan A = 15-------tanA, tan2 A - lOtanA + 9= 0, =>, tan A = 1 or tan A = 9, _ ., ., 14 1, f, => tan B and tanC are 2, 3 or —, -, respectively, 3 3, ‘, => AABC is always on acute angled triangle and sum of all, possible values of tan A is 10., , • Ex. 59. (m + 2) sin 0 + (2m -1) cos 0 = 2m +1, if, , 3, (a)tan0=±, , 4, (b)tan0 = |, (d)tan0=_^_, , (c) tan0 =, (m - 1), , (d)tan 0 =, , 4a2 - y2, cos a cos P = —-—=4x2 +y, , and, , Similarly, we can write (1) as a quadratic in sin 0, giving, two values sin a and sin p, such that, sin a + sin P =, , x‘ +y, , sin a sin P =, , and, , 4a2 - x2, , x2 +y2 ', , • Ex. 61. Lety = sin2 x+cos4 x. Then, for all real x, (a) the maximum value of y is 2, 3, (b) the minimum value of y is —, , (m + 1), , Sol. (b, c) The given relation can be written as, (m + 2) tan 0 + (2m -1) = (2m + 1) sec 0, =>(m + 2)2 tan2 0 + 2(m + 2)(2m - 1) tan 0 + (2m - I)2, , (c) y<i, , (d) yi-l, 4, , www.jeebooks.in, Sol. (b, c)y = cos4 x - cos2 x + 1, , = (2m + I)2 (1 + tan2 0), , => [(m + 22) - (2m + I)2] tan2 0 + 2(m + 2) tan 0 + (2m - l)2, - (2m + I)2 = 0, , 3, 2, 1, cos x — + 2I 4

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www.jeebooks.in, 66, , Textbook of Trigonometry, , 3, ,, 1, = — and y is maximum when cos2 x —, 2, maximum, 3 =1, = _1 + _, , ■ ’T min, , 4, , 2, , . 1+n, (c)-----1-n, , is then, , I2 + m2, 2n, , Tt, , (d)a + 3 = — if I = m, , 4, , Sol. (a, b, c, d) Now, I2 = sin2 a + sin2 3 + 2 sin a sin 3 and, m2 = cos2 a + cos2 3 r 2 cos a cos 3, , • Ex. 62. If in fsABC, tan A + tan B + tan C = 6 and, , Sol. (b, c) tan A + tan B + tanC = 6, => tan A tan B tan C = 6, 2 tan C = 6, tan C = 3, tan22 C _ 9 _ 9, sin2C =, 1 + tan2 C 1+9 10, , (by subtracting), , 2 cos2a + cos23) = m2 - I2, , (b) 8 :5: 9, (d) 5:8 :5, , (a) 8 : 9:5, (c) 5: 9:5, , (by adding), , 2cos(a-3) = /2 + zn2-2, , tan A tan B = 2, then sin2 A : sin2 B: sin2 C is, , - m — Ii2, 2 cos(a + 3) cos(a - 3) + 2 cos(a + 3) =, , cos(a + 3) =, , —(i), , m2-l2, m2 + l2, , • Ex. 65. Let f(x) =ab sin x + b^l - a2 cos x + c, where, , | a | < 1, b > 0 then, (a) maximum value of /(x) if b is c = 0, (b) difference of maximum and minimum values of/(x)is, 2b, (c) /(x) = c if x = - cos-1 a, , From Eq. (i), tan A + tan B = 3 and tan A tan B = 2, tan A - tan B, = ± -J{(tan A + tan B)2 - 4 tan A tanB)}, , (d) /(x) = c if x = cos"1 a, , we get, tan A = 2,1 and tan B= 1,2, 11, A, ■ 2- n, 1, 4, . 2 ., 4, sin A =------ -,------ and sin B =------ ,, 1+4 1+1, 1+11+4, , =>, , Sol. (a, b, c) f(x) = ab sin x + b^jl - a 2 cos x + c, where, | a | < 1, b < 0, /(x) = ^a2b2 + b2 - b2a2 sin(x + a) + c, , • 2 & 8 5, , . 2n, 5 8, sin A=—, — and sin B = —,—, 10 10, 10 10, sin2 A:sin2B:sin2C = 8:5:9 or 5:8:9, , _ b^l - a'2, , = b sin(x +a) + c, where tana, , ab, ab, = b cos(x - a) + c, where tan a =, byjl - a 2, , • Ex. 63. lf0<x,y <180° anJsin(x -y) = cos(x+y) = —,, 2, then the values ofx andy are given by, (a) x = 45°, y = 15°, (c) x = 165°, y = 15°, , a, , ■Jl — a',2, , f(x)^ ~ f(x)min=c + b-(c-b) = 2b, , (b) x = 45°, y = 135°, (d) x = 165°, y = 135°, , or, , Sol. (a, d) sin(x - y) = - => x - y = 30° or 150° (1), cos(x + 1) = - => x + y = 60° or 300° (2), 2, Since x and y lie between 0’ and 180°, (1) and (2) are, simultaneously true when x = 45°, y = 15°, or x = 165°,, y= 135°. But, for the values given by (b) or (c), (1) and (2) do, not hold simultaneously., and, , • Ex. 64. If sin a + sin p = I, cos a cos P = m and, , tan, , a, 2, , (RA, . . ., tan - = zX* 1), then, I 2J, , (a) cos(a - 3) =, , f2 + m2 - 2, , f(x) = cifx+a=0, x = -a or x=-cos-1 a, , • Ex. 66. If(x - a) cos 0 + y sin 0, = (x - a) cos (J) + y sin <|) = a and tan, , - tan, , 2>, , £| = 2b,, 2J, , then, A, , 1, , (a) y2 = 2ax - (1 - b2)x2, , (b) tan — = — (y + bx), 2 x, , (c) y2 = 2bx - (1 - a!2)X2, , (d) tan — = — (y - bx), 2 x, , Sol. (a, b) Let, tan | | = a and tan f — j = 3, so that a - 3 = 2b., \2/, I 2J, , e, , www.jeebooks.in, 2, , (b) cos(a + 3) = —2—lm +i, , Also,, , 1 - tan2, 2, cos 0 =---------e, 1 + tan2, 2, , 1 -a2, 1 + a2

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , And, , 2 tan -I, 2), sin 0 =--------(0, 1 + tan2, , 1-022, , 2a, 1 + a2, , . . A, , Also, from a + 0 = — and a - 0 = 2b, we get, x, a = — + b and 0 = — - b, x, x, 0, 1, /, tan - = — (y + bx), 2 x, 1 (y - bx), tan -0 = —, 2 x, , 20, , Similarly, cos 0 = ----- - and sin 0<p == —, 1 + 02, 1 + 02, , Therefore, we have from the given relations, f 2a ', 1-a2, =a, (x-a), t, +, t, j + a2,, 1+a, xa2 - 2ya + 2a - x = 0, , =>, , x02 - 2y0 + 2a = 0, , Similarly, , We see that a and 0 are roots of the equation, xz2 - 2yz + 2a - x = 0,, a + 0 = — and a0 = ——, x, x, Now, from (a + 0)2 =(a - 0)2 + 4a0 , we get, , So that, , ST =(2il)>+, XJ, , =>, , 67, , and, , 3, • £x. 67. //cos(0 - y) + cos(y - a) + cos(a - 0) = — then, 2, (a) Z cos a = 0, (b) £ sin a = 0, (c) Z cos a sin a = 0, (d) Z (cos a + sin a) = 0, Sol. (a, b, d) The given expression can be written as, 2 [cos 0 cos y + cos y cos a + cos a cos 0], + 2 [sin 0 sin y + sin y sin a + sin a sin 0], + (sin2 a + cos2 a) + (sin2 0 + cos2 0), + (sin2 Y +cos2 Y) = 0, => (cos a + cos 0 + cos y)2 + (sin a + sin 0 + sin y)2 - 0, , X, , y2 = 2ax - (1 - &z)x2, , =>, =>, , Zcos a = Oand Zsin a = 0, Z(cosa +sina) = 0, , JEE Type Solved Examples:, Statement I and II Type Questions, • Ex. 68. Statement I tan 50 - tan 30 - tan 20, = tan 50 tan 30 tan 20., , Statement II x = y + z, => tan x - tan y - tan z = tan x tan y tan z, (a) A, (b)B, (c)C, (d)D, So/. (a) v 50 = 30 + 26, tan 30 + tan 20, =>, tan 50 = tan(30 + 20) =, 1 - tan 30 tan 20, => tan 50 - tan 50 tan 36 tan 20 = tan 30 + tan 20, =>, tan 50 - tan 30 - tan 20 = tan 50 tan 30 tan 20, , • Ex. 70. Statement I Ifa, b,cE R and not all equal, then, _ (be + ca + ab), sec 0 =-----------------,, (a2+b2+c2), Statement 11 sec 0 < -1 and sec 0 > 1, (a) A, (b)B, (c)C, (d)D, Sol. (d) v a2 + b2 + c2 - ab - be - ca, , =>, , or, • Ex. 69. Statement I The maximum value of, sin 0 +cos0 is 2., Statement II The maximum value of sin 0 is 1 and that of, cos 0 is also 1., (a) A, (b)B, (c)C, (d)D, Sol. (d) v sin 0 + cos 0 <^2, , = - {(a-b)2 +(b - c)2 + (c - a)2} > 0, 2, a2 + b2 + c2 > ab + be + ca, ab + be + ca, a2 + b2 + c2 <, , sec 0 < 1, which is false., • Ex. 71. Statement I II (1 + sec 2' 0) = tan 2n0 cot 0, r=1, , n, , Statement II FI cos(2r~' 0) =, r=1, , sin(2n0), , 2" sin0, , www.jeebooks.in, Maximum value of sin 0 + cos 0 is 72, , But maximum value of sin 0 is 1 and that of cos 0 is also 1, which is always true., , (a) A, (c)C, , (b)B, (d)D

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www.jeebooks.in, 68, , Textbook of Trigonometry, , „, FI (1 + cos 2" 0), Sol. (a) v FI (1 + sec 2r 0) = ^4------------Fl cos 2" 0, , .’. cos’, , 471, 2n ), a + cos a + — + cos 31 a + —, I, 3 J, 3, , (, 47C, [, 2ti, = 3 cos a cos a + — cos a + —, k, 3, I, 3, , fl 2 cos2 (2r”'0), r»l, , • Ex. 74. Statement I sin 2 > sin 3, , FI cos(2r0), , r=l, , I ft, , 1, , Statement II Ifx, ye — ,7t \,x<y, then sin x > sin y, , 2" • fl cos(2r-10) fl cos(2r-1 0), , J, , 12, , So/. (a), , -cos(2,e) fl cos(2"'0), , y-axis, , COS0, , 2* sin(2"0), • cos 0, 2" sin 0, cos (2”0), , sin 2, sin 3, , = tan(2n0) • cot 0, , 2, , 2, , 3, , X-axis, , • Ex. 72. Statement I cos 36° > sin 36°, ft, , • Ex. 75. Leta, P, y > 0 anda. + 3 + y = —, 2, , Statement II cos 36° > tan 36°, (a) A, , (b)B, , (c)C, , (d)D, , qi, /?!, al, Statement I tan a tan P “ ~ + tan Ptan Y “ ~, , 6, , 71, , Sol. (b) Since, cos 0 > sin 0 for 0 < 0 < —, , c*, , + tan y tan a - — <0, where n! = 1.2, , So, Statement I is true., Now,, cos 36° > tan 36°, o sin 36°, =>, cos 36° >-------cos 36°, , tan P tan y, tan y tan a are in AP., Statement II tan a tan P + tan P tan y + tan y tan a =1, _, , 1 + cos 72° > 2 sin36° = 2 sin(30° + 6°), 1 + 2 sin 9° cos 9° > cos 6° + 2 cos 30° sin 6°, which is true, , (, , 271, , I, , 3, , =>, E tan a tan P = 1, /. Statement II is true., , • Ex. 73. Statement I cos3 a + cos3 a + —, , + cos3, , Statement I tan a tan P =, , J—, , Statement II Ifa +b + c = 0 <=>a3 +b3 +c3 =3abc, (b)B, , (d) D, , _ , . ., , (, , 271, , (, , 471, , I, , 3., , Sol. (a) •/ cos a + cos I a + — + cos a 4----„, , z, , 'ft, , = cos a + 2 cos(a + 7t) cos —, , bl, tanP tan y = —, , 471, , a +—, 3, , (l _ . 271 I], [|, 47t, = 3 cos a cos a + — cos a + —, , (c)C, , 7t, , Sol. (d) Statement II a + p =---- y, 2, tan a + tan P, 1, 1 - tan a tan p tan y, , cos2 36° > sin 36°, , (a) A, , n, then tan a tan P,, , 3, , c!, tan a tanB = —, 3, a! bl c!, —+—+—=1, 6 2, 3, =>, al = 1 bl = 1 c! = 1, => tan a tanP, tan y tan a and tan p tany are not in AP., /.Statement I is false., Hence, (d) is the correct answer., , and, , • Ex. 76. Statement I The triangle so obtained is an equi­, lateral triangle., , www.jeebooks.in, = cos a +(- 2 cos a), , 0, , Statement II If roots of the equation be tan A, tan B and, tan C, then tan A + tan B + tan C = 3^3

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 69, , y, , Sol. (b) tan A + tan B + tan C = 3-^, , J— y=x2+x+1, , tan A tan B tan C = 3, tan A + tan B+ tanC, tan A tan B tan C, =s triangle does not exists., and, , y= sin x, ----- -X, , • Ex. 77. Let us define the function f(x) = x2 + x + 1, Y, , Statement I The equation sin x- /(x) has no solution., Statement II The curve y =sin x andy = f(x) do not inter­, sect each other when graph is observed., , Sol. (a) Let y = sin x and y = x2 + x + 1, , . . ., .. ., (, 1Y 3, Since, - 1 < sin x < 1 and y = ( x + -i- Y + ' < 2) 4, It is clear from the graph that no two curves intersect each, other., , JEE Type Solved Examples:, Passage Based Questions, Passage I, (Ex. Nos. 78 to 80), , or, , k >0’, , D <0, , Consider, /(x) = (x + 2a) (x + a - 4) (a G R),, g(x) = k(x2 + x) + 3k + x(fc e R) and, , n, , Here, D = (k +1)2 - 4k. 3(k + 1) < 0, , h(x) = (1 - sin0)xz + 2(1 - sin0)x - 3sin0, (, , kx2 +(k + l)x + (3k + 3) > 0 V x, , k2 +2k + l-12k2 — 12 <0, =>, , IT, , 10 G/?-(4n +1)—,ne 11, , life2 + lOfc — 1 > 0, , (Ar 4- l)(llfc — 1) > 0, k<-l, , • Ex. 78. If f(x) < Ofor -1 < x < 1, then ‘a ’ satisfies, (aU<a<3, 2, , (c) -3 < a < - 2, , (b)-^<a<^, 2, 2, 1, (d) - 3 < a < 2, , (.*. fc>0), , k> —, , =>, , 11, , • Ex. 80. If the quadratic equation h(x) =0 has both roots, , complex, then 0 belongs to, , Sol. (a) Given, f(x) = (x + 2a) (x + a - 4), = x2 + (3a - 4) x + 2a (a - 4)., , f(-l)<0', n, /(l)<0, , or k>11, , 1, , wf°.Ti, A 2J, , (a) "T-T, 2 2, , f7t 7ny, , o’*, , ,..(7n Tin, , d IT’T, , (c) PT, , On solving, we get - < a < 3, 2, • Ex. 79. If g(x) > - 3 for all real x, then the values ofk, , are given by, , Sol. (d) Given, (1 - sin0)x2 + 2(1 - sin0)x - 3sin0 = 0 has both, roots complex, then D < 0, (1 - sinG) (1 + 2sin6) < 0, (sinO - l)(2sin0 + 1) > 0, (-) ve number, , (a)-i<^<J-, , (b) - 1 < k < 0, , (c) 0 < k <, , (d)K-, , =>, , 2sin0 + 1 < 0, , sin0 < - 2, , 11, , =>, , 06, , 77t llTt'), , www.jeebooks.in, Sol. (d) g(x) = k (x2 + x) + 3k + x > - 3 V x, , =>, , k(x2 + x) + 3k + x + 3>0V x, , 6 ’, , 6

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www.jeebooks.in, 70, , Textbook of Trigonometry, , • Ex. 83. Sum of all values ofx satisfying the equation, , Passage II, , (Ex. Nos. 81 to 83), , Let /(0) = sin0 - cosz0 - 1, where 0 g R and m < /(0) < M., , 1, , x=, , +Jlll+, , • Ex. 81. Let N denotes the number of solution of the, , equation f(Q) =0 in [0,4n] then the value of, 1 ', log^M + log^, is equal to, , N + 1,, , (b)I, , <4, Sol. (d) x =, , (b)1, , 2, , =>, , 2j, , I, , 4, , /, . \2, = | sin0 + - |, 2j, , 9, , f(0L = 0-’4, , -9, , 9xz - 12x + 3x —4 =0, =>, , 4, , (3x — 4) (3x + 1) = 0, 4, -1, x = — and x = — (rejected), 3, 3, , 4, , Passage III, (Ex. Nos. 84 to 88), , k, , 9, , 9, , ’=0, , 4, , 4, , 4, , 2/, , M=0, -9, , Hence, m = —, M = 0, 4, Now, /(0) = O, =>, (sin0 + 2) (sin0 - 1) = 0, =>, sin0 = 1, =>, , 9x2-9x-4=0, , -9, m=—, 4, , /(6U, , 4, , x=~ + x, 9, 9x2 = 4 + 9x, , = sin20 +sin0 -2, = | sin0 + - |, , oo, , 14, -+x, 9, , x=, , Sol. (c)/(0) = sin0 - (1 - sin20) - 1, , =>, , 4, -+, 9, , 14, '4, -+ -+, 9, 9, , (d)-1, , (c,T, , <», is, , o, , 71, , J 571, , 0 = — and —, 2, 2, Hence, N = 2, i.e. number of solution s of sin0 = 1 in [0, 4 k],, 1 ', , log^W + log^, , On the basis of above information answer the following, questions., • Ex. 84. If x sin3 0 +y cos33 0 = sin0cos0 and, , x sin 0 - y cos 0=0 then (x, y) lie one, , (b) a parabola, (d) a hyperbola, , (a) a circle, (c) an ellipse, , Sol. (a) We have, x sin3 0 + y cos’ 0 = sin 0 cos 0, and, , x sin 0 - y cos 0=0, , -.(i), , ...(H), , N + l,, , hMS, , = 10^o.|, , The method of eliminating ‘0’ from two given equations, involving trigonometrical functions of ‘0’. By using given, equations involving ‘0’ and trigonometrical identities, we shall, obtain an equation not involving ‘0’., , From Eq. (ii), tan 0 = —, , -1, 2 ’, , y, , • Ex. 82. The value of(4m +13) is equal to, X, , (a) 0, (c) 5, , (b)4, (d) 6, , sin 0 =, , y, , J(x2+y2), , and cos 0 =, , www.jeebooks.in, -9, Sol. (b) As m = —, so (4m + 13) = 4, 4, , )

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www.jeebooks.in, 72, , Textbook of Trigonometry, , sin 0 cos 0=0, ab, Adding Eqs. (iii) and (iv), we get, ( x2 - a2 sin2 6, cos2 0, +-------a, b, a, I, , sin2 0, , ...(iv), , -, , “b~, , or, , b, , COS2 0, +------=o, , x2 - a<2, a, , a, , y2 - b2, b, , 2, , = 0 or, , a, , 2, , b, , 7, , J EE Type Solved Examples:, Matching Type Questions, • Ex. 89. Match the statement of Column I with values of, Column II., , Column-I, , Column-II, , (A) The number of real roots of the equation, cos’ x + sin4 x = 1 in (- n, n) is, , IpH, , 7b> The value of V3 cosec 20° - sec 20° is, , (q) 4, , (C) 4 cos 36° - 4 cos 72° + 4 sin 18° cos 36°, equals, (D) The number of values of x where, xe[- 2n, 2n], which satisfy, cosec x = 1 + cot x, , (s) 2, , 2, , • Ex. 90. Match the statement of Column I with values of, Column II., Column-II, , (A) The number of solutions of the equation, , (B) If sin 6 + sin 0 = - and cos 0 + cos 0 = 2,, 2, then cotf ? +, , (s), , (p), , No solution, , (q), , 1, 3, , < 2 J, , cos x = 0 or cos5 x = 1 +sin2 x, J, , r, , cos x = 0 => x---- , —; cos s x = 1 + sin x, 2 2, x=0, [•/ LHS < 1 and RHS > 1], , . 2 a + sm---H a sin|----71 a I, (C) sin, <3, (D) If tan 0 = 3 tan 0, then maximum value of, tan2(0 - 0) is, , Sol. (A) -> (r), (B) -> (s), (C) -> (p), (D), , 71 „ 7t, , x = ~—,0, —., 2, 2, (B) Vi cosec 20° - sec 20°, , (r), , 1, , (s), , 4, , (q), , (A) | cot x | = cot x + ——, sinx, , 1, , Vi cos 20° - sin 20°, , cos 20°, , sin 20° cos 20°, , V3, 1, 2 — cos 20° - - sin 20°, , 2, , I2, , 4 sin 40°, L ----------=4, , sin 40°, , sin 20° cos 20°, , (C) 4 cos 36° - 4 cos 72° + 4 sin 18° • cos 36°, \, / r, V5 -1, =4, -4, +4, 4 7, 4, , =3, (D) cosec x = 1 + cot x;, , 2, , | cot x | = cot x + —-— (0 < x < rt), sin x, , cos7 x = (1 + sin2 x) cos2 x, , V3, sin 20°, , 371 7C, X =------ , —, , Column-I, , Sol. (A) -» (r), (B) -» (q), (C) -> (r), (D) ->, (A) cos' x+ sin4 x = 1, , It 7t, , n, , V X - — 6 - 271 - —, 2tt - 4, 4, 4, , 1, , sin x + cos x, , sin x, sin x + cos x - 1 and sin x 0, 1, 71, cos x----4, 2, , sin x, , If 0 < x < — => cot x > 0, 2, 1, So, cot x = cot x +, sin x, , 1, , = 0 no solution, , sin x, , 1, It — < cot X < 7t, - cot X = cot X +, 2, sin x, 2 cos x, + —— = 0, sin x, sin x, 271, , 1 + 2 cos x =0 and x * 0 => x = —, 3, (B) Since, sin $ + sin 0 = - and cos 0 + cos <|> = 2 has no, 2, solution., (C) sm a + sm---- a • sin — + a13, 3, , www.jeebooks.in, 7t, 71 7t, X----- = - 27t + —, —, , 4, , 4 4, , .2, , • 2 TC, , ,2, , 3, , = sin a + sm — sin a = —, 3, 4

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www.jeebooks.in, 73, , Chap 01 Trigonometric Functions and Identities, , (D) tan0 = 3 tan 0, tan 0 - tan 0, 2 tan 0, tan(0 - <{)) =, 1 + tan 0 tan 0 1+3 tan2 0, 2, - ------------------ Max of tan 0 > 0, cot 0 + 3 tan 0, cot 0 + 3 tan 0 >, (using AM > GM), 2, =>, , Column-II, , IpH, , (C), , 0 1, , If x = y cos — = z cos —, then, 3, 3, , 13, , 2, , — and minimum value occurs at cos x = 1 and it is, 4, 13, /.The required ratio is —., • Ex. 92. Match the statement of Column I with values of, Column II, , If a, fl, y and 8 are four solutions of the, equation tanl 0 + — | = 3 tan 30, no two of, k, 4), which have equal tangents, then the value, of tan a + tan fl + tan Y + tan 8 is, , (s) I, , cos(0) + 02), , 25, , Sol. (A) -> (s), (B) -> (p), (C) -> (q), (D) -> (t), (A) Given, tan a = 3 and tan fl = 2, 3-2, tan a - tan P, tan(a - P) =, 1-tan a tan fl 1 + 3x2, , Ifsec(a-fl),seca andsec(a + fl) are in, , (D) T,, , ,,, ., , 41t, , 2ft, , . ., , ., , (C) We have, x = y cos — = z cos — = k (say), 2, 3, 4n, 2lt, cos —, cos —, 1, 1, 1 1, ___ 3_, — = —9 — = ___ 3_, =>, z, k, x k y, k, 4n, 2n, 1 1 1 if, — + - + - = -| 1 + cos — + cos--3, 3, x y z k, , -1, , (s), , 0, , a, , 7, , = tan a • tan — a = tan a cot a = 1, , 2 cos fl -1, 2 - cos fl, , If cos a =-------- — (0 < a < fl < n), then, , .£, , I n, 1, (B) We have, tan a -tan(2n - 1) a = tan a • tanl —---- 1 la, , /.The given expression = 1., , (r), , A.P. (with fl * 0), then cos a sec - =, 2, , = 2 X-7= X-7= =—, <50, <50 25, , J, , (q) V3, , cos(0j - 0, ), , tan 0, tan 6; tan 0, tan 0< =__________, , 1, 7, sin(a - P) = -7= and cos(a - P) = -7=, V50, <50, sin 2(a - P) = 2 sin(a - P) cos(a - P), n, 1, 7, 7, , U, , Column-II, , Column-I, , 4, , 1, , +3, , 2, , (B) If co5(9,..- 0.7.) + cos(9^ 9J = 0then, , 1 7t, , 4, , .’. Maximum value occurs at cos x = - - and it is 1, , 2, , (D) The ratio of the greatest value of, 2 - cos x + sin2 x to its least value is, , _1, , COS X + —, , 4, , (q) o, , xy + yz + zx =, , /., , 1, , 1, , cos x + 2, , 13, , • Ex. 91. Match the statement of Column I with values of, Column II., (A) The tangents of two acute angles are, 3 and 2. The sine of twice their, difference is, (B), If n =—, then tan a tan2a tan 3a..., 4a, tan(2n - l)ais equal to, , = - (cos2x + cos x) + 3 = -, , 2, , (cot 0 + 3 tan (|>)2 > 12 => tan2(0 - 0) <, , Column-I, , (D) We have, 2 - cos x + sin2 x =2 - cos x + 1 - cos2 x, , tan 2, ---- g- is equal to, tan2, , Sol. (A) -> (s), (B) -+ (r), (C) -> (p), (D) -> (q), .,,_ _ ., (n it 'I 1 + tan 0, (A) Usmg tan| 0 + — I = -------- -, , and, , 3 tan 30 =, , 3(3 tan 0 - tan3 0), 1 - 3 tan2 0, , the given equation becomes, 3 tan4 0 - 6 tan2 0 + 8 tan 0 - 1 = 0, , If tan a, tan p, tan y and tan 8 are the roots of this equation,, then the sum of these roots, tan a + tan P + tan y + tan 8, equals zero, since the coefficient of tan3 0 is zero., (B) The given equation can be written as, cos 0, cos 0, + sin 0; sin 02, =>, cos 0] cos02 -sin0, sin02, + cos 03 cos 04 - sin 03 sin 04, cos 03 cos 04 + sin 03 sin 04, , www.jeebooks.in, £ 1-1, , 1, , 2, , 2, , k, , => xy + xz + yz = 0, , =0, , 1 + tan 0, tan 0,, l-tanO^an©,, , 1 - tan 03 tan 04 _ Q, l + tan03tan04

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www.jeebooks.in, 74, , Textbook of Trigonometry, , 2 + 2 tan 9, tan 02 tan 0, tan 04, , =>, , (1- tan 0( tan02)(l + tan 0, tan 04), , cos P, 1 - cos2 — = 1------------ ^-52, l + 2sinz2, 2, 2P, 2 P, 1 + 2 sin — - cos, 2, 2, , =o, , Showing that tan 0, tan 02 tan 0, tan 04 = - 1., (C) For the given A.P., we have, 2 sec a = sec(a - P) + sec(a + P), which gives, 2, _, 1, [, 1, cos a, , l + 2sinz 2, l-sin22, 2 P, 1 + 2 sin2 2, 2, , cos(a + p), , cos(a - P), , 2 cos a cos p, cos2 a - sin2 P, cos2 a - sin2 P = cos2 a cos P, , 1 + 2 sin2 —, 2, 3 sin2 2, . 2 a, 2, sin —, 2 P, 2, 1 + 2 sin —, 2, Dividing equation (ii) by (i), we get, ,a „, 2 P, tan — = 3 tan —, 2, 2, a, tan —, , cos2 a(l - cos p) = sin2 P, , . 2 P, fl = 4. sin, — cos 2P, , cos2 a 2 sin 2, , ,, , 2J, , 2, , 2, , ,P, , cos a sec* — = 2, 2, 2 cos P - 1, (D) 1 + cos a = 1 +, 2 - cos P, 2 - cos p + 2 cos P - 1 _ 1 + cos P, , 2 - cos P, 2 a, , =>, , cos —, 2, , 2 - cos P, , 2 P, cos —, 2, 1 + 2 sin2 2, 2, , (ii), , __ 1=73, P, tan —, ...(i), , 2, , J EE Type Solved Examples:, Single Integer Answer Type Questions, • Ex. 93. tan 46° tan 14° - tan 74° tan 14° + tan 74° tan 46°, is equal to, , Sol. (3), , tan 46° + tan 14° = tan(46° + 14°) = 73, 1 - tan 46° tan 14°, tan 74° - tan 14°, , 1 + tan 74° tan 14°, , • Ex. 94. Maximum value of the expression, log3(9 - 2cos2 0-4 sec2 0) is equal to, Sol. (1) For the expression a cos20 + bsec2© if b > a, then, minimum value attains at cos2 9 =sec20 = 1, => max of {9 -(2cos20 + 4sec20)} = 3, , = tan(74°-14°), , So, maximum of log3(9 - 2 cos2 0 + 4sec20)) = 1, .-(ii), , 3, I, 71 i, • Ex. 95. Let x g I 0, — I and log 24,inx(24cosx)=-, then, , tan 74° 4- tan 46°, = tan(74° + 46°), 1 - tan 74° tan 46°, 3, , ...(iii), , Sol. (9) (24 sin x)3/2 = 24 cos x, , From Eqs. (i), (ii) and (iii), , tan 46° tan 14° = 1 tan 74° tan 14° =, , find the value of cosec2x., , tan 46° + tan 14°, , 73, , tan 74° - tan 14°, , 73, , 724 (sinx)3/2 = cosx, , 24sin3 x = cos2 x = 1 - sin2 x, -1, , Put sin x = t, we get, 24t3 + t2-l = 0, , www.jeebooks.in, tan 74° tan 46° = 1 -, , tan 74° + tan 46°, , -73, , tan46°tanl4o-tan740tanl40+tan740tan46° = 3, , (3t-l)(8t2+3t + l) = 0, , >0

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 1, t=3, 1. ., 1, t = - i.e. sin x = 3, 3, cosecx = 3, cosec2x = 9, , =>, , =>, , 75, , 9 3, n, 5n, 37t 1, = - + - cos, + cos -— + cos, 8 8, 7, 7, 7 J, sin^, 1 useS = —~~ cos 91 + °2, 2, 2, sin2, 9 3 21, =-+— = —, 8 16 16, a-b = 21-16 = 5., , • Ex. 96. Ifx and y are non-zero real numbers satisfying, , xy(x2 -y2) = x2 +y2, then find the minimum value of, • Ex. 98. In any triangle, if, , x1 + y2., , (sin A + sin B + sin C) (sin A + sin B - sin C) = 3 sin A sin B,, C, then the angle — (in degree)., , Sol. (4) Put x = r cos0 and y = r sin0, , Hence, we have to minimise r2?, Now, r2 cos0 sin0r2(cos20 - sin20) = r2, , Sol. (6) We have, (sin A + sin B)2 - sin2 C = 3sin A sin B, , r2 sin20 cos20 = 2, sin 40, , 4, , sin2 A - sin2 C + sin2 B = sin A sin B, sin( A + C) sin( A - C) + sin2 B = sin A sin B, , =1, , sinB[sin(A - C) + sin(A + C)] = sin A sinB, [using, sin (A + C) = sin B], (sin B * 0), 2 sin A cos C = sin A, C, „ 1, cos C = - =>, — =6, 2, 10, 10, , sin 40, , r2 = 4 cosec2 40, r2min — 4, , • Ex. 99. Find the exact value of the. expression, • Ex. 97. Using the identity, , 3 1, 1, sin4 x = — - cos 2x + - cos 4x or otherwise, if the value of, , 2, , 8, sin4, , 71, , 8, 371, , + sin4, , 7, , + sin4, , 1 ), , 571, , a, , 7, , b, , where a and b are, , 1, , coprime, find the value of{a - b)., , \7, and, , sin, , or, , sin, , 4 3n, l7, , or, , sin, , I7, , 45n, , 3, , 1, , 37C, , 8, , 2, , 7, , = - + -cos, , I7, , ...(ii), , 7t, 1, — cos, , (iii), , 7, , 8, , On adding Eqs. (i), (ii) and (iii), we get, 371, / 571, , sin4— | + sin 4|, , l7, , I7J, , + sin, , 8, , 2, , 7, , 8, , ' 7, , 1, , cos 40° cos 20°, , + 1 + 2 cos20°, , 2cos 40°cos20°, 2 sin 10° sin 20°, , +1 + 2cos20°, sin80°, 2 sin 10° 2sin 10° cos 10°, + 2 cos20° +1, cos 10°, , = 2(1 - cos20°) + 2cos20° + 1 = 3, 2sin20° COS209, = 2-2-sinl0°-cos20°, Alternatively T, =, cos 10°, , = 2(sin30° - sin 10°), , I7, , 3tt, 57t, 1, 9, 1, = - + -cos, —cos, , + 2(cos60° + cos 20°), , cos20°, , = 2sin3C-siniy+1+2cos2tf,, , 20tt, 1, 1071, 3 1, + - cos, ------ cos, 7, 7, 8, 8 2, , 4|, , 1, , 1 cos 20° - cos 40°, 2, , 571, 1, 71, 3 1, — cos, = - + - cos, 8, 7, 8 2, 7, , 4 | 371, , Similarly, sin, , ...(i), , 1271, 1, 671, 3 1, + - cos, ------ cos, 7, 7, 8, 8 2, , I7, , 1, , 2 cos 40°, , 471, 1, 2n, 3 1, + -cos, ------ cos, 7, 7, 8, 8 2, , Sol. (5) sin4f—, , sin 40° sin 80° sin 20°, T =--------+-------------------sin 80° sin 20° sin 40°, 1, 1, + 4 cos 40° • cos20° Sol. (3) We have,, 2 cos20°, 2 cos 40°, , ■, , 71, , + - cos, 2, 7, , 5tc, 1, — cos, , 8, , 7, , +, , 3n, 1, 71, 1, — cos, -cos, 7, 8, 7, 2, , T, = 1 - 2 sin 10°, sin 80° 2sin40° cos 40°, t2 = sin 20° ”, sin 20°, = 4 cos 20°-cos 40°, T, = 2[cos60° + cos 20° ] = 1 + 2 cos2(F, , www.jeebooks.in

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www.jeebooks.in, 1, , Chap 01 Trigonometric Functions and Identities, , tan, , tan, , • Ex. 106. If sin x, + sin x2 + sin x3 + ... + sin x,M, = 2008,, , 3n, , 350 -1, , then find the value of sin2008 x, + sin2008 x2i + sin2008, , 7C, , 3”-l,, , Sol. (2008) We know that, sin x( < 1V i, =>sin x, +sin x2 + sin x, +.... + sin x20M < 2008, Thus, equality holds only when each of the terms is 1, i.e., sin x( = 1V i = 1, 2, 3,..., 2008., and consequently, cosx( =0,Vi = 1,23,. 2008, Now, sin200* x, + sin2008 x2 + sin2008 , +... sin zoos X20M, , n, ct = ——, 3" -1, ,, tan3a, k =------tana, 8, 1 - 3 tan2 a =, 3k-1, , So,, , i, , *3, , + ... + sin2008 x.2008', , Let,, , S, , 77, , a = 8, b = 3, a-b = 5, , = 1 + 1 + 1 + ... + 1 = 2008, , • Ex. 104. If sec A tan B + tan A sec B = 91, then the value, , • Ex. 107. If 4 sin 27° = 7a + TP, then the value of, , of(sec A sec B + tan A tan B)2 is equal to, , (a + P - aP + 2)4 must be, , Sol. (8282) (sec A sec B + tan A tanB)2, , Sol. (400), We know (cos 27° +sin 27° )2, = 1 + sin 54 = 1 + cos 36°, =>, cos 27° + sin 27° = 7(1 + cos 36°), , =, , - (sec A tanB + tan A sec B)2, 1 +sin Asin B, , 2, , cos A cos B, , sin B + sin A, , 2, , cos A cos B, , [v LHS > 0], , Also, cos 27° - sin27° = 7(1 - cos 36°), , [v cos 27° > sin 27°], 2 sin 27° = 7(1 + cos 36°) - 7(1 - cos 36°), , _ 1 + sin2 A sin2 B - sin2 B - sin2 A, , cos2 A cos2 B, _ 1 - sin2 B cos2 A -sin2 A, , 1+, , cos2 A cos2B, cos2 A cos2 B — j,, _----------cos2 A cos2 B, , '5 4-1, , 4, , 1-, , 4, , 4 sin27° = yj(5 + 75) - ^(3 - 75), , => (sec A sec B + tan A tan B)2 = (91)2 + 1 = 8282., , On comparing, we get, a=5 + 75,p=3-7s, a+p = 8,ap = 10-2^5, , • Ex. 105. //(25)2 + a2 + 50a cos0, = (31)2 + b2 + 62 6cos0 =1 and, , a + P - aP + 2 = 2 V5, , 775 + ab + (31a + 25b) cos 0=0, then the value ofcosec2 0 is, , (a + P-aP + 2)4 = 400, , Sol. (1586) We can write (a + 25 cos 0)2 + (25)2 - (25 cos 0)2 = 1, jr, , and, =>, , (a+25 cos 0)2 = 1 - (25 sin 0)2, , Similarly, , (b + 31cos 0)2 = l-(31sin 0)2, , Multiplying we get, [(a + 25 cos 0) (b + 31 cos 0)]2 = [1 - (25 sin 0)2], , [1 - (31 sin0)2], =>, , [ab + (31a + 25b) cos 0 + 775 cos2 0]2, , = 1 - (625 + 961) sin2 0 + (775 sin2 0) 2, =>, , (- 775 + 775 cos2 0)2 = 1 - 1586sin2 0 + (775 sin2 0)2, , ♦ Ex. 108. lfQ<A< — and sin A +cos A + tan A, 2, + cot A + sec A + cosec A = 7 and sin A and cos A are the, roots of the equation Ax2 -3x 4-a =0, then the value of 25a, , must be, Sol. (28) sin A and cos A are the roots of the equation, 4x2 - 3x + a = 0, then, , sin A + cos A = —, sin A cos A = —, 4, 4, , (i), , Also,sin A + cos A +tanA + cot A + sec A + cosec A = 7, , cosec2 0 = 1586, , www.jeebooks.in

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www.jeebooks.in, 78, , Textbook of Trigonometry, , (sin A + cos A) +, , sin A cos A, +------kcos A sin A, , +, , .•.The given expression =, , 1, , 1, +------ =7, ^cos A sin A, , 1 -, , sin 20°, , 1, , Ji cos 120°, , 1, , cos 60°, , sin 20°, , sin 60° cos 20°, , sin 60° cos 20° - cos 60° sin 20°, (sin A + cos A) +, , 1, + (sin A + cos A), sin A cos A, sin A cos A, , sin 20° cos 20° sin 60°, , sin(60° - 20°), , 3 4 3 „, -+-+-=7, , =>, , 4, , a, , sin 40°, , a, , 2, , V3, , _x, , =X, , ?=, , i, , =A, , I, , =X, , x—, 2, , 3 7 „, -+ - = 7, a, , 4, , =>, , V3, , ’=7-2, , 25, , 4, , 4, , a, , A2 = —, 3, , =>, , ,,, 256, 16, Then, 9A + 81A +97=9x — + 81 x — + 97, , 25a = 28, 2 sin 0, , • Ex. 109. Given that, f(n&) =, , 9, 3, = 256 + 432 + 97 = 785, , and, , cos 20 - cos 4n0, , sin A0, , /(0) + /(20) + /(30) + ... + /(n0) =, , , then the, , sin 0 sin p0, , value of\i - A is, , Sol. (1) f (n0) =, , • Ex. 111. //Iog,0 sin x + log10 cos x = -1 and, log,0(sin x +cos x) = - °^10 n\—1, ■, then the value of ‘n/3’ is, 2, , 2 sin 20, cos 20 - cos 4n0, , sin 2x, = -l, 2, sin 2x, , Sol. (4) Given, logl0, , 2 sin 20, 2 sin(2n + 1)0 sin(2n - 1)0, , or, , 2, , sin((2n + 1)0 -(2n - 1)0), sin(2n + 1)0 sin(2n - 1)0, , •, , or, , n, , 10, 1, , sin 2x = 5, , sin(2n + 1)0 cos(2n - 1)0, , - cos(2n + 1)0 sin(2n - 1)0, sin(2n + 1)0 sin(2n - 1)0, , = cot(2n - 1)0 - cot(2n + 1)0, , n, , Also, , log10(sin x + cos x) =, , or, , lQgJ\10, 2, , log10(sin x + cos x)2 = log10, , .-.f(0) + /(20)+/(30)+... + f(n0), , = cot 0 - cot(2n + 1)0, , or, , 1 + sin 2x = —, 10, 1, n, 6, 1 + - = — or, 5 10, 5, , sin(2n + 1)0 cos 0 - cos(2n + 1)0 sin 0, , sin(2n + 1)0 sin 0, sin 2n0, , or, or, , sin(2n + 1)0 sin 0, , n = 2n and p = 2n +1, Hence, p - A = 2n +1 - 2n = 1, , n, , 10, , n, 10, , —=4, 3, , • Ex. 112. //498[16cos x +12 sin x] = 2k +60, then the, , maximum value ofk is, , 1, , 1, , cos 290°, , Ji sin 250'|O, , • Ex. 110. If -------- +, , - = A, then the value of, , 9A4 +81A2 + 97 must be, , Sol. (785) Here, cos 290° = cos(270° + 20°) = sin 20° and, sin 250° = sin(270° - 20°) = - cos 20°, , Sol. (4950) 16 cos x + 12 sin x = -J162 + 122 cos(x - a), a, = tan-1, , 4J, , 12k + 601 < 498 x 20 as | cos(x - a) | < 1, k < 4950, , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , ,C, (C\, i A-B, C, 2sin---- 2sin — cos, -1 + -=0, 2, 2, 2, 2, , =>, , C, A4sin, . 2C A4 sin, . fCA, |, — cos, 2, , A-B, , 2, , 2, , + 1=0, , A-B, cos2, >1, 2, , A-B, , which is only possible if cos2, , 2, =>, , A-B, , cos2, , 2 J, A-B, , 2, , 3P - 1, , > 0, using number line rule., , ...(i), , Now, Eq. (i) is quadratic in (sin C/2) and is real., D>0, 2 A-B, A-B, =>, - 16 >0 => cos2, 16 cos, - l>0, 2, 2, , =>, , P -3, , =>, , =1, , 81, , T, , 1/3, , 3, , which shows k < 1 / 3 or k > 3, • Ex. 121. Let A, B, C be three angles such that A = 7i/4 and, , tan B tan C = p. Find all possible values of p such that A, B, C, are the angles of triangles., Sol. Let us assume AABC., .’., A + B + C =7t, 371, , =1, , Also,, =>, , =0, , B + C = 7t-- =, •••(>) [vA = 7t/4, given], 4, 4, 0< B.C <3ti/4, tan B tan C = p, sin B sin C _ p, , cosB-cosC 1, sinB-sinC + cosB-cosC _ 1 + p, , =>, A=B, (ii), Similarly, we can show B = C, C = A. Hence, the triangle is, equilateral., , cosB-cosC -sinB sinC, , 1-p, , cos(B - C) _ 1 + p, 2P, ,, •, irtan3A . ,, . sin3A, • Ex. 120. If--------- = k, show that--------- =------ and, tan A, sin A, P-1, , cos(B + C), , =>, , cos(B - C), , hence or otherwise prove that either k > 3 or k < -., , Jl + P, J"P, , 1-p, , 3ti, , cos, , 4J, [using Eq. (i), B + C = 3n/ 4], , tan 3 A _ 3 tan A - tan3 A, , Sol., , tan A, =>, , 3-tan2 A, ---------l-3tan=, 2A, , => cos(B-C) =, , —= k, tan A, , l-3tan2A, , =>, , (3fc - l)tan2 A = k - 3, , =>, , tan2 A =, , =>, , =>, , fc-3^, , 3sinA - 4sin3 A, , sin A, , sin A, 4, , =>, , _4____, 3fc -1-), 1+, P -3 j, , 3------, , 3-, , —7=- < cos(B — C) < 1, , 72, , 72(p -1), 1+p, , __ 1, , 72 4i{p -1), o<i + ^-ii, , =>, , 2, , ...(iii), , 72, , From Eqs. (ii) and (iii), we get, 11 + p, <1, , = 3-4sin2A, , = 3-------l + cotzA, cosec A, , = 3-----1+, , =>, , ...(i), , 3k -! J, , sin 3 A, , 3-4sin 2 A = 3-, , ••■(ii), , Since, B or C can vary from 0 to 3n/4., 0<(B-C)<37t/4, , ,, , (3 - tan2 A) = P(1 - 3tan2 A), , Now,, , 1+P, , 72(p - 1), , and ^(P-1)S1, , (p + l)-72(p-l), <0, 72(p - 1), , and, , P-1, , 4, , 2p, —>0, (P-1), , 1, tan2 A, , 0, , [using Eq. (i)], , and, , 1, , [p-(^ + D2], , (p-i), -U, 1, , >0, , (V2+1)2, , p < 0 or p > 1 and p < 1 or p > (72 + 1).2, , 4(P-3)_3P-3-P + 3_ 2k, 4(P-1)~, P-1, ~P-1, , ...(ii), , The combining above expressions;, p <0or p >(72 + I)2, , www.jeebooks.in, Again from Eq. (i),, - 2., P-3, tan A =------3P-1, , i.e., , [tan A * 0 and tan2 A > 0], , pe(-~,0)u[(72 + l)2,~).

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www.jeebooks.in, 82, , Textbook of Trigonometry, , • Ex. 122. If ABC is a triangle and tan, , ABC, tan —, tan — are, 2, 2, 2, , D, , 1 + tan A tan B, , in HP, then find minimum value ofcot —., 2, Sol., A+ B+C=n, A B 7t, C, —+—=—, =>, 2 2, 2, 2, (A B^, n C, cot — + — = cot, I 2 2J, 2, 2, =>, , A, B, cot —cot---- 1, 2, 2, A, B, cot — + cot—, 2, 2, , 1, tan(A - B), , cot(A - B) =, , Now,, , tan A - tan B, 1+----- y = 1 + 1 = RHS, x, x y, , 1, (ii)2cos A = x + —, since 4sin2 A = 4, x, 2, , 1, - 4 cos2 A = 4 - X + —, X, , co,(l), , A, B, C, A, B, C, cot — • cot — • cot — = cot — + cot — + cot —, 2 2, 2, 2, 2, 2, t,, A, B, C, . Tm, But tan — , tan—,tan — are in HP, 2, 2, 2, ABC, =>, cot—, cot—,cot— are in AP, 2 2, 2, A, C, B, cot— + cot— = 2cot —, 2, 2, 2, From Eqs. (i) and (ii), we get, A, B, C n, B, cot—cot —cot— = 3cot —, 2, 2, 2, 2, , (i), , 4sin2 A = i2, , or, , 1, x----x, , l, , 2, , ■, , 2sin A = i( x - — ], , xJ, , \, (ii), , Similarly,, , 2cos B = y + —, y, , „2sm, . BD = 1/y---I, , (ii), , yj, , Now, 2cos(A - B) = 2 [cos A cos B + sin Asin B], , A cot —, c =3, cot —, 2, 2, As we know, AM > GM, A, C, cot — + cot —, A, C, 2_____ 2_> 1 cot —, cot —, 2, 2, 2, =>, , 2 cot —, ____ 2 > 3, 2, , 2, , I ~4, , 4 sin2 A = -, , (iii), , 2, 4, , 1, 2, , [{using Eq. (iii)], , cot— > y/3, 2, .•.Minimum value of cot — is y/l., 2, , i, , ;2, , X-----.X, , y), , I, , 1, y-_, y)j, , 1, 1, y x, xy + — + — + — - xy +------ y _£, xy x y, xy x y, , 1 2y+2^, = — + — = RHS, 2 . x, y. y x, , Ex. 124. //tan 0 tan 0 =, , a -b, , then prove that, , a+b, (a - bcos 2 0)(a - bcos 2 0) is independent ofB and 0., , Sol. Let us put,, Ex 123. (i) If tan A - tan B = x and cot B - cot A = y., , tan0 = tj and tan 0 = t2, 22, , Prove that cot(A - B) = — + —., x y, , a-b, , (ii) If 2cos A = x + — , 2cos B = y + — , then show that, x, y, , 2cos(A - B) = — + —., Y *, tan A - tan B, Sol. (i) If cot B - cot A = y => ----------------- = y, tan A tan B, x, — = tan A tan B, y, , (i), , =----- a+b, , given, tan 0 tan 0 =, , Also, cos 20 =, , l-tan20 _ 1-t2, l + tan20, , 1 + t2, , 1 - tan2 0, , 1 - t2, , l + tan20, , l + t2, , a-b, , a+b, ■(ii), , www.jeebooks.in, COS20 =, , Now, (a - frcos20).(a - I>cos20), , •(iii)

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www.jeebooks.in, 85, , Chap 01 Trigonometric Functions and Identities, , = 2sin, , =>, , X, (—, 3a ., 1 |, —+ n+- a, 2, 2J, , f, , sin, , 2, , cosC{sin AsinB - cosC}, , 1 ., 3a, n + - a-----2J, 2, 2, , 7 \, a, (n +2) a, (n-l)a, 2sin —•$ = 2sin, •sin, 2, 2, 2, (n-l)a, sin, (n +2) a, 2, •sin, S=—, . a, 2, sm —, 2, (n - l)a, sin, • 2% n, 2, •sin— = 0, a, 2, sin —, 2, , • Ex. 131. Sum the series fl + cos a +, , sin A sin Bsin C, [v cosC = cos(n - (A + B)) = - cos(A + B)], cosC{sin AsinB + cos(A + B)}, , 7, , +cos 2a, , sin A sin Bsin C, cos C {sin A sin B -t- cos A cos B - sin A sin B}, sin A sin Bsin C, cos A cos B cos C, ,, „, „, --------------------- = cot A • cot B • cot C = RHS, sin A sin Bsin C, , • Ex. 133. In A.ABC, ifcotB = cot A + cot B+cotC,, prove that sin3 0 = sin(A - 0)sin(B -0)sin(C -0)., So/. We have, cot0 = cot A + cotB + cotC, =>, cot(0) - cot(A) = cot B + cotC, cos0 cos A —--------cosB4.--------cosC, , sin0 sin, sinAA---- sinB---- sinC, cos0sinA - cosAsin0 _ cosBsinC + sinBcosC, , + 71 +cos3a +... ton terms., , Sol. We have,, fl + cos a + fl + cos2a + fl + cos3a +... + fl + cosna, = y2cos 2a 12 + x/2cos22 a + J2cos2— + ... ton terms, V, 2, r, a, 2a, 3a, = v2{ cos— + cos— + cos— + ... + to n terms ■, !, 2, 2, 2, . na, sin —, 4, , ------ — • COS, , a, , <\a, , /, , . a, sm —, , {using formula}, , 4, . na, sin—, (•, a, 4 .cos{(n + 1)—, = fl----- —, a, .al, 44, sin —, 1, 4, , • Ex. 132. If A + B + C = it, show that, cot A + cot B + cot C - cosec A cosec B ■ cosec C =, cot A • cot B • cot C, So/. LHS =, +, ------------ 1---------sin A, sinB, sinC sin A-sinB-sinC, cosA-sinBsinC + cosBsinAsinC + cosCsinAsinB- 1, , sin Asin Bsin C, sinC(cosAsinB + cosBsinA) + cosCsinAsinB - 1, , sin A sin Bsin C, sinCsin(A + B) + cosCsin AsinB - 1, , sin A sin 0, , sin Bsin C, , sin(A -0) =, , Similarly,, , sin(B - 0) =, , and, , sin(C-0) =, , sin2 Asin0, , (i), , sin Bsin C, sin2 Bsin0, , -(ii), , sin A sin C, , sin2 Csin0, , (iii), , sin Asin B, Multiplying Eqs. (i), (ii) and (iii), we get, sin(A - 0)sin(B -0)sin(C -0) = sin30., • Ex. 134. If A, B, C and D are angles of a quadrilateral, , , . A . B . C . D 1 ,, and sin — sin — sin — sin — = —, then prove that, 2, 22, 2 4, , A = B=C = D=n/2., C I IT, (n • A . B^ (, C . D\, C, D, Sol. Now, 2sin —-sm— • 2sm —-sm— = 1, I, 2, 2, 2, 2), , => {cos^ A-B, , (a+bMI (c-d\, (C+ D, - cos ------ }{cos ------- - cos ------\ 2 Jjl \ 2 J, k 2, , =1, , Since, A + B = 2n - (C + D\ the above equation becomes,, (, A-B, A + B}(, C-D, A+B, cos------- cos-------- cos--------+ cos, =1, , 2, , I, , => cos2!, , sin A sin Bsin C, sin2 C +cosCsinAsinB-1, , sin A sin Bsin C, [usingsin(A + B) = sin(n - C) = sinC], , sin Bsin C, sin(B + C), , sin 0 sin A, sin(A - 0), , 2, , A+B, 2, , - cos, , - CO!, , A-B, 2, , A, , 2, , A-B, 2, , A+B, 2, cos, , 2, , C-D, , - CO!, , C-D, , +1, , = 0., , 2, , This is a quadratic equation in cos, , A+B, , which has real, , www.jeebooks.in, cos C.sin A sin B - cos2 C, , sin A sin Bsin C, , roots., , 2

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www.jeebooks.in, 88, , Textbook of Trigonometry, , Sol. Let A = B, then 2A + C = 180°, and, 2tan A + tanC = k, Now,, 2A + C = 180°, tan 2 A = - tanC, Also,, 2tan A + tanC = k, 2tanA + tan(180 - 2A) = k, 2tanA - tan2A = k, 2 tan A, 2 tan A---- — = kF, 1 - tan2 A, , fc2>| < o, y, , k/3, , ...(ii), , =>, , /'(x) = 6x2 — 2kx = Q, , x = k/3,0, Following cases arises, (i) k < 0, three graphs of cubic equation (ii) are possible., Clearly, in all these case, only one triangle is possible, and the condition for that triangle to be possible is, /(0) < 0 => k < 0 so for k < 0 only one isosceles triangle, is possible., , W3, , 0,, , X, , Oj, , r, I 27 J, , k 1-— >0, , (iii) k = 0, graph will be shown as, so no such triangle is, possible. Hence, the solution for mentioned, conditions;, .*. (A) either k < 0 or k = 3^3, (B) k >3^3, y, , y, , y, , ■x, , In figure (iii), no such triangle is possible. The, condition is f(kf3) > 0, , 2tan3 A -fctan2 A + k = 0, tan A = x, x > 0 (as A < 90°), /(x) = 2x3 - kx2 +k, , x, , In figure (ii), one such triangle is possible. The, condition is f(k/3) = 0, =>, k = 3^3., , 2tanA(l - tan2 A - 1) = k - fctan2 A, , Let,, Then let,, , => k > 3>/3, , ■X, , ■X, , (ii) k > 0, three graphs of the cubic equation (ii) are, possible. In fig. (i), two such triangle are possible. The, condition is /(/t/3) < 0., , o, , *x, , (C) Clearly, there will never exists three or more than, three non-similar isosceles triangle for any value, of k., , www.jeebooks.in

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www.jeebooks.in, Trigonometric Functions and Identities Exercise 1:, , K“ Single Option Correct Type Questions, 1. The value of V fsin, , 2nn, , - cos----, , (a) 2, (c)0, , (a), , 2. Given, a2 +2a + cosec2 — (a + x) = 0, then which of, 2, J, , (d), , (n - l)cos A, sin A, , sin A, (n + l)cosA, , ,^ocos(3'*‘0), , 10. The value of, (cos4 l° + cos4 2° + cos4 3° + ... +cos4 179°)-, , 3. The minimum value of the function, , (sin4 1° + sin4 2° + sin4 3° +... + sin4 179°)equals to, , /(x) = (3sin x - 4 cos x -10)(3sin x + 4 cos x - 10), is, 195 - 60V2, (a) 49, (b), 2, (c) 84, (d) 48, 8, 1, 4. The value of expression V--------------------- equal is to, 0 = O1 + tan(lO0)°, , (b)^, 4, , (a) 5, , 14, (C)^, 3, , 6. If tana, tan 0 are the roots of the equation, , sin2(a + P) + psin(a + 0) cos(a + 0) +</cos2(a + 0) is, , (a) independent of p but dependent on q, (b) independent of q but dependent on p, (c) independent of both p and q, (d) dependent on both p and q, , cos(x - y\ is, ,, , '3fl2“2, , 9a2 -2, , 2, , ^(cos x + cos 2x + cos 3x)2 + (sin x + sin 2x + sin 3x)2,, , (a)l + 2cosx, (c) 1 - 2cosx, , (b)l+sin2x, (d) None of these, , ------------------- ;--------------------is equal to — (where, p and, 5sec2 0 - tan2 0 + 4cosec20-----------------q, , (a) 14, (c) 16, , 14. Let/,(□) =, 7t, , (b) 15, (d) 18, sina +sin3a +sin5a + ... + sin(2n - l)a, , cosa + cos 3a + cos 5a +... + cos(2n - l)a, , ( ji A, Then, the value of fA — is equal to, \ 32 J, , COS, , 2*“, , ... cos, , 7a—, 2 -2, (b)^, 2, 5q2 -2, (d), 2, , (a)—, , q are coprime), then the value of (p + q) is, , 7. The value of the product, , 2^, , sin x + sin y = a and cos x + cos y = 2a The value of, , 13. If the maximum value of the expression, , x2 + px + q = 0, then the value of, , it, , 11. Suppose that ‘a is a non-zero real number for which, , then P(x)is equal to, , (b)-l, (d) 1, , 22009, , (b) — 1, (d) 0, , 12. LetP(x) =, , 5. The value of -Jl -sin2110° -sec 110° is equal to, , 71, , (a) 2 cos 1°, (c) 2sin 1°, , (c), , 9, (d);, 2, , (a) 2, (c)-2, , (b)P = 3Q, (d)3P = Q, , (a)P=2Q, (c)2P = Q, , (b) a = —1; —G/, 2, (c) a G R; x G 0, (d) a, x are finite but not possible to find, , cos, , sin A, , (n -1) cosA, , (b), , 9. IfP = (tan(3" + 10)-tan0)andQ = £, , (a) a = 1; — G 1, 2, , 71, , (c), , sin A, (1 - n)cosA, , V J?g9))then, , the following holds good?, , COS, , , then tan(A + B) equals to, , 1 - n cos' A, , (b)l, (d)-l, , . ( Tt, sin ------ cos, l 22°09 I, , 8.IftanB^sinAcosA, , 11, , 11, , .Tk, , is equal to, , 71, , 23, , WpU, , ( n ] ., cos — , is, , <22 J, , (a) -Ji + 1, , (b)^-l, , (C)2 + >/3, , (d)2-V3, , www.jeebooks.in, 15. The minimum value of sin x + cos x +, , cos x + sin x, , cos4 x - sin4 x, , (a) 2, , (c)72, , (d)l, , is

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www.jeebooks.in, 90, , Textbook of Trigonometry, , 16. If a = cos(2012 7t), b = sec(2013 rt) and c = tan(2014 n),, , then, (a) a < b < c, , (b) b < c < a, , (c) c < b < a, , (d) a = b < c, , 17. In a AABC, the minimum value of, 2 A, , 2 B ,, , sec — + sec — + sec, 2, 2, (a) 3, (c) 5, , 2 C ., , 23. One side of a rectangular piece of paper is 6 cm, the, adjacent sides being longer than 6 cm. One corner of the, paper is folded so that is sets on the opposite longer side., If the length of the crease is I cm and it makes an angle 9, with the long side as shown, then I is, , ■, , —is equal to, 2, (b) 4, (d) 6, , 18. The number of ordered pairs (x, y) of real numbers, , satisfying 4x2 - 4x + 2 = sin2 y and x2 + y2 < 3, is equal, , to, (a) 0, (c)4, , (b) 2, (d) 8, , (b)60°, (d)150°, , 20. An equilateral triangle has side length 8. The area of the, region containing all points outside the triangle but not, more than 3 units from a point on the triangle is :, (a) 9(8 + n), (b) 8(9+7t), (c) 9f 8 + — |, k, 2J, (d) 8|9 + -l, k, 2J, , 21. If a cos3 a + 3a cos a sin2 a = m and, , asin3 a + 3acos2 a sina = n. Then,, (m + n)2'3 + (m - n)2'3 is equal to, , (a) 2a2, (c)2a2'3, , 6, sin2 0 cos0, 3, (d), sin2 0, , 3, sin0 cos2 0, 3, (c), sin 0 cos 0, , (b), , (a), , 19. In a AABC, 3sin A + 4 cos B = 6 and 3 cos A + 4sin B = 1,, then ZC can be, (a) 30°, (c)90°, , 6— *, , (b)2a1/s, (d)2a3, , 22. As shown in the figure,AD is the altitude on BC and AD, produced meets the circumcircle of AABC at P where, DP = x. Similarly, EQ = y and FR = z. If a, b,c respectively, denotes the sides BC, CA and AB, then — + — + — has, 2x 2y 2z, , the value equal to, , 24. The average of the numbers n sin n° for n = 2, 4,6,... 180, , (a) 1, , (b) cot 1°, , (c)tanl°, , (d)2, , 25. A circle is inscribed inside a regular pentagon and, another circle is circumscribed about this pentagon., Similarly, a circle is inscribed in a regular heptagon and, another circumscribed about the heptagon. The area of, the regions between the two circles in two cases are >1], and A2, respectively. If each polygon has a side length of, 2 units, then which one of the following is true ?, 5, , 25, , (a)A=-^, , (b)A=-\, , (c) A = — A, , (d)A=A, , 49, , 7, 49, , 18, , cos 2 (5r)°, where x° denotes the x, , 26. The value of, r=1, , degrees, is equal to, , (a)0, (c) —, , 27. Minimum value of 4x2 - 4x | sin x | - cos 2 0 is equal to, , (a)-2, (c)-|, , (b)-l, (d)0, , 28. If in a triangle ABC, cos 3A + cos 3B + cos 3C = 1, then, , one angle must be exactly equal to, (a)j, , (b)y, , (c)n, , 29. If | tan A | < 1 and | A | is acute, then, 7(1 + sin 2A) + 7(1 - sin 2A), , is equal to, , www.jeebooks.in, (a) tan A + tanB + tanC, (b) cot A + cos B + cot C, (c) cos A + cosB + cosC, (d) cosec A + cosecB + cosec C, , 7(1 + sin 2A) ~7(1 - sin 2A), , (a) tan A, (c) cot A, , (b) - tan A, (d) - cot A

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 30. For any real 0, the maximum value of, cos2 (cos 0) + sin2 (sin 0) is, , (b) 1 +sin2 1, (d) does not exist, , (a) 1, (c) 1 + cos2 1, , 31 Minimum value of27co,2x -8r, •81 ,n2x is, (a)-5, , (b)|, , (c)—, 243, , (d)27, , 32. ABCD is a trapezium, such that AB and CD are parallel, BC1 CD. If Z.ADB = 0, BC = p and CD = q, then AB is, equal to, p2 + <72 cos 0, (a), + <?2)sin 0, (b)I —- -------- -------------pcos 0 + q sin 0, p cos 0 + <7 sin 0, !, (p2 + <7" ) sin 0, p, (c), (d), (p cos 0 + q sin 0)2, p2 cos 0 + q2 sin 0, , 33. If 4na = 71, then cot a cot 2a cot 3a... cot(2n - l)a is, equal to, (a)0, (c) n, , 39. f(0) = | sin 0 | +1 cos 0 |, 0 g R then, , (b)/(0) e [0, V2], (d) _f(0) g [1, V2], , (a)/(0)e(O.2j, (c)/(0)e[0,1], , 40. If A = cos(cos x) + sin(cos x) the least and greatest value, , of A are, (a) 0 and 2, (c) - <2 and 72, , (b) - 1 and 1, (d) 0 and -Ji, , 41. If Un = sin zi0 sec" 0, Vr = cos n0 sec" 0^1, then, -|, IL/ n., lt/, -----------—+, +---------- -is, - is equal to, U ., n Vn, , (a)0, , (b) tan 0, , tan n0, (c) - tan 0 +, n, , (d) tan 0 +, , 7C, , 42. If 0 < x < — then range of /(x), , 3, , + se<, , is, , (b) 1, (d) None of these, , (a) IkV3, -7=.00 ., , (b), , 34. If in a triangle ABC(sin A + sin B + sin C), , 4 ', (c) ( 0,4, , (d), , (sin A + sin B - sin C) = 3 sin A sin B, then angle C is, equal to, (a) 30’, (b) 45°, (c) 60°, (d) 75’, , 35. If a, P, y are acute angles and cos 0 = S‘--—, sin a, siny, cos <p =----- - and cos(0 — 0) = sin P sin y, then, sin a, tan2 a - tan2 P - tan2 y is equal to, , (b)0, (d) None of these, , (a)-l, (c) 1, , n sin a cos a, L ., D. ., .., 36. If tan P = -------------—, then tan(a - P) is equal to, 1 - n sin2 a, (b) (1 - n) tan a, (d) None of these, , (a) n tan a, (c) (1 + n) tan a, cos 0, , sin 0 ., , 37. If ------ =------- , then, a, , b, , (a) a, , a, , b, +----- ------ is equal to, sec 20 cosec 20, (b)b, , (a) X > 1, (c) 1 < 2a < 3, , I, , y/3j, , (b) 0 < A < 1, (d) None of these, , 44. The expression 3] sin4, , 371, •• 4/„, ----- a + sin (3tc + a), , 2, , J, , - 2^ sin6 — + a + sin6 (5n - a) is equal to, <2, J, (a)0, (c) 1, , (b)-l, (d)3, , 71 1, 1, n, \, it I, 45. The maximum value of sin | x + — | + cos | x + — in the, 6, 6J, I 7t I, interval 0, — is attained at, , I 2J, , < \ n, , (b)£6, , (a)u, , (c)£3, , (d)^2 ‘, , 46. If cot2 x = cot(x -y) • cot(x - z\ then cot 2x is equal to, -L 71, , (, , x*±k, , is, (a) a straight line passing through (0, - sin2 0) with slope 2, (b) a straight line passing through (0, 0), (c) a parabola with vertex (1, - sin2 1), ( Jt, , A, (d) a straight line passing through the point I, - sin" 11 and, , 4, , 3’", , 43. If A = sin8 0 + cos14 0, then for all values of0,, , (d) a + b, 38. The graph of the function cos x cos(x + 2) - cos2(x + 1), , 91, , (a), , 4/, (tan y + tan z), , (c) - (siny + sinz), 2, , (b) - (cot y + cot z), 2, (d) None of these, , 47. The minimum value of the expression, sin a + sin P + sin y, where a, P, y are real numbers, satisfying a + P + y = 71, is, (a) positive, (b) zero, (c) negative, (d) None of these, , www.jeebooks.in, parallel to the X-axis

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www.jeebooks.in, 92, , Textbook of Trigonometry, , x, 48. If cos x - sin a cot P sin x = cos a, then tan — is equal to, , .., a, (a) cot —, , 2, , p, tan -, , /(c)\ tan —a tan -P, 2, , complete set of values of x is, n 3n', (a) x g [ 0, —, 4, 2 4;, 3tt, 71 71, (b)xe, —, ,k, 4’ 2, 4, , tan y, , (b) cot, , 2, , (d) None of these, , 2, , 51. If x G (0,71) and sin x • cos33 x > cos x • sin3 x, then, , 49. If cos4 0 sec2 a, - and sin4 0 cosec2 a are in AP, then, 2, , cos8 0sec6 a, - and sin* 0 -cosec6a are in, 2, (a) AP, (b) GP, (c) HP, (d) None of these, , . ., , c xe, , | 71 7t, , \4 2, (d) None of the above, 52. If u = -ja22 cos2 0 + 52 sin2 0 + -Ja2 sin 2 0 + b2 cos2 6, , 50. The maximum value of, cos a | • cos a j • cos a, ■... • cos a n under the restriction, , 0<a,,a,......a, < — and cot a. - cot a, -...-cota =1, ’ 2, 2, , then the difference between the maximum and, minimum values of u2 is given by, (a) 2(a2 + b2), (b) 2y]a2 + b2, (c)(a + b)2, , (d)(a-b)2, , 53. For a positive integer n, let fn (0) =, , is, , o»±, , (a)4-, , Mi)-, , (d)l, , (c)f, , (1 + sec 0), , 2, , (1 + sec 20)... (1 + sec 2"0), then, , 2’, , 22, , tan 0, , M£)=, , 0, , -1, , g Trigonometric Functions and Identities Exercise 2:, More than One Option Correct Type Questions, 54. Suppose cos x = 0 and cos(x + z) = -. Then, the possible, 2, value(s) of z is (are)., , 2, , (b)T, 6, (d)HE, on, , oo, , (a) xyz = xz + y, (b) xyz = xy + z, (c) xyz = x + y + z, (d) xyz =yz + x, , n, , 55. Let fn (0) = 2sin - sin — + 2sin —, 2, 2, 2, ■, , 50, , „ ., , 0, , ., , 70, , O •, , 0 • /n, , Ji, , Mv)-, , «(”)■ 0, n e N, , 0, , /, , H0, , sin— + 2sm — sm— + ... + 2sm— sin(2n + 1)—, n G N,, 2, 2, 2, 2, 2, then which of the following is/are correct?, i, , J/in2"©,, n -0, , n =0, , 6, , ‘ n, , n=0, , z = £cos 20 Osin2" 0, then, , I A 71, , (a)7, 7n, T, , cos 2n0,y=, , 57. For 0 < 0 < —, if x =, , 1, , Ji, , 56. Let P = sin 25° sin 35° sin 60° sin 85° and, , Q = sin 20° sin 40° sin 75° sin 80°. Which of the following, relation(s) is (are) correct ?, (a) P + Q = 0, (b)P-Q = 0, (c) P2 + Q2 = 1, (d) P2 - Q2 = 0, , 58. Let P(x) = cot2 x, , , \, 1 + tan x + tan x, 2, , 1 + cot X + cot X J, , +, , ' cos x - cos 3x + sin 3x - sin x, , . Then, which of the, 2(sin2x + cos2x), , following is (are) correct?, (a) The value of P(18°) + P(72°) is 2., (b) The value of P(18°) + P(72°) is 3., , (c) The value of, , +, , is 3., , www.jeebooks.in, (d) The value of, , +, , is 2.

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 59. It is known that sin 3 = - and 0 < 0 < n, then thes value, , 67. If 3sin P = sin(2a + p), then, , (a) [cot a + cot(a + P)] [cot P - 3 cot(2a + p)] = 6, (b) sin p = cos(a + p) sin a, (c) 2 sin p = sin(a + P) cos a, (d) tan(a + P) = 2 tan a, , 2, 73 sin(a + p) - ----- — cos(a + p), 1171, COS-----, , 6_______, , of, , is, , sin a, (a) independent of a for all p in (0,7t), , (b), , 68. Let Pn(u) be a polynomial is u of degree n. Then, for, every positive integer n, sin 2nx is expressible is, (a) P^(sin x), (b) P2„(cos x), (c) cos x P^.^sinx), (d) sin xP^/cos x), , for tanp > 0, , 73, 73(7 + 24 cot a), , for tanP < 0, 15, (d) zero for tan p > 0, (c), , r., n sin a - cos a ,, 69. If tan 9 =----------------- , then, sin a + cos a, (a) sin a - cos a = ± 7i sin 0, , 3, 60. In cyclic quadrilateral ABCD, if cot A = - and, 4, —12, tan B = —then which of the following is (are) correct?, 12, , 93, , 16, , (b) sin a + cos a = ± 72 cos 0, , (c) cos 26 = sin 2a, (d) sin 20 + cos 2a = 0, 70. If cos 59 = a cos 9 + b cos3 9 + c cos5 9 + d, then, , (a)sin£) = —, , (b) sin(A + B) =—, , (c) cosD = ——, 13, , (d) sin(C + D) - ——, 65, , (a) a =20, (b) b = -20, (c) c = 16, (d) d =5____, , 61. If the equation 2 cos2 x + cos x - a = 0 has solutions,, , then a can be, (a)21, , (b)^, , then x2 = a2 + b2 + 2y]p(a2 +b2)- p2, where p is equal, , (c)2, , (d)5, , to, (a) a2 cos2 a + b2 sin2 a, (b) a2 sin2 a + b2 cos2 a, (c) 1 [a2 + b2 +(a2 - b2) cos 2a], , 71. x = a2 cos 2 a + b2 sin2 a = -ja2 sin2 a + b2 cos2 a, , 62. If A = sin 44° + cos 44°, B = sin 45° + cos 45° and, C - sin 46° + cos 46°. Then, correct option(s) is/are, (a) A <B<C, (b) C < B < A, (c) B > A, (d) A = C, , (d) 1 [a2 + b2 -(a2 - b2) cos 2a], , 63. If tan(2a + p) = x & tan(a + 2P) = y, then [tan3(a + P)]., z, , [tan(a - P)] is equal to (wherever defined), , 1+xy, , >, , (c), , x 2 + y2 ., <, , ,, , 72., , ^2,,2, , (d), , sin A - sin B ?, , (a) 2 tan", , (b>x=i4, , +, , sin A + sin B, , A-B, , (n, even or odd) is, , (d)xy + x- y + l = 0, , 1 + cos, , A-B j, , (b) 2 cot", , 2, , (c)0, , 2 J, , (d) None of these, It, , I, , (2Jt + l)jt, , 4k, , w, , (a)P(3) = l, , 1 + cos, , (2Jfc -1)71, 4k, , 1 + cos, , (4k - l)7t, , (b)75 -2, , (c) (9-475) (2+75), , (d) (9+ 475) (2-75), (then one of the value of y is, , (c)P(5) =, , 3-75, , Then, , 4k, , 2-72, (b)P(4) = —, 16, , 16, , (a)2-75, , Sn, , xn, , cos A - cos Bz, , 73. Let P(k) = 1 + cos —, I, 4k, , y+l, , l JC I, 65. If tan — = cosec x - sin x, then tan 2 — I is equal to, , 66 If y = i, , xn, , equal to, , x2-y2, 1 - x2y2, , 64. If x=sec 0 - tan 0 and y = cosec 0 + cot 0, then, , (a)x = ^-i, y-i, i+X, (c)y=-----1-x, , cos A + cos B, , (d)P(6) =, , 2-73, , 32, , 16, , 74. If x= a cos’ 9 sin2 9, y = asin3 9 cos2 9 and, , (x2+y2f, , www.jeebooks.in, (xy)’, , 71+sin4A-l, , (a) - tan A, , (c) tan^~ + j, , (b) cot A, (d) - cot f, , (p, q e N) is independent of 9, then, (a) p = 4, (b) p = 5, (c)q=4, (d)g=5, , +A, , >5*.

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www.jeebooks.in, 94, , Textbook of Trigonometry, , g Trigonometric Functions and Identities Exercise 3:, Statement I and II Type Questions, ■ This section contains 11 questions. Each question contains, Statement I (Assertion) and Statement II (Reason)., Each question has 4 choices (a), (b), (c) and (d) out of, which only one is correct. The choices are, (a) Both Statement I and Statement II are individually true, and R is the correct explanation of Statement I., (b) Both Statement I and Statement II are individually true but, Statement II is not the correct explanation of Statement I., (c) Statement I is true but Statement II is false., (d) Statement I is false but Statement II is true., 75. Statement I tan a + 2 tan 2a + 4 tan 4a + 8 tan 8a, + 16 cot 16a = cot a, , Statement II cot a - tan a = 2 cot 2a, , 76. StatementIIfxy + yz + zx=l, then, S—-—=, , 2, , Statement II In a AABC sin 2A +sin 2B - sin 2C, = 4 cos A cos B sin C, 77. Statement I If a and 0 are two distinct solutions of the, , equation a cos x + b sin x = c, then tan, , a +0, , is, , 2, , Statement II Solution of a cos x + b sin x = c is possible,, if—, + b2) < c < -^(a2 + b2), , 78. Statement I If A is obtuse angle in AABC, then, tan B tan C > 1., tan B + tan C, Statement II In AABC, tan A =, tan B tan C - 1, 47t, , 7, , Statement II cos — + i sin, , Statement II The curve y =sin x or y = cos x intersects, the X-axis at infinitely many points., 81. Statement I The numbers sin 18° and - sin 54° are the, roots of a quadratic equation with integer coefficients., , Statement II If x = 18°, cos 3x = sin 2x and if y = - 54°, sin 2y = cos 3y., 82. Statement I The minimum value of the expression, sin a + sin 0 +sin y where a, 0, y are real numbers such, that a + 0 + y = 7tis negative., Statement II If a + 0 + y = 7t, then a, 0, y are the angles, of a triangle., , 271 ., 7, , + sin, , 8n, 7, , _______, , _______, , g, , 83. Statement I If 2 sin — = ^/1 + sin 0 + .y 1 - sin 8 then V<2J, 2J ’, 2, , lies between 2nn + — and 2n7i + —., 4, 4, , Statement II If — < 0 < — then sin— > 0., 4, 4, 2, f, , 7T |, , I, , 2), , 84. Statement I If 2 cos 0+ sin 0=10#— then the value, , independent of c., , ) + s*n, , - 71 < X < 7t., , /a \, , (1 + x2) 7n(i + x2), , 79. Statement I sin, , COJ2 X, , 80. Statement I The curve 7 = 81“" x + 81co> x - 30, intersects X-axis at eight points in the region, , _1, 2, , is complex 7th root of, , of 7 cos 0 + 6 sin 8 is 2., 1, , 7t, , Statement II If cos 20 - sin 0 = -, 0 < 8 < —, then, 2, 2, sin 0 + cos 60 = 0., , 85. Statement I If A > 0, B > 0 and A + B = —, then the, 3, , maximum value of tan A tan B is -., 3, , Statement II If a, + a2 + a3 +... + an =k (constant),, then the value axa2a3...an is greatest when, ai =a2 =a3 =- = an, , unity., , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 95, , § Trigonometric Functions and Identities Exercise 4:, Passage Based Questions, Passage I, , (a) 8x5, (b) 8xJ, (c) 8x5, (d) 8x’, , (Q. Nos. 86 and 87), If a, b, c are the sides of A ABC such that, 32?, +, =Qthen, , 92. The value of sec — + sec — + sec — is, 7, 7, 7, (a) 4, (b)-4, (c)3, (d)-3, , 86. Triangle ABC is, (a) equilateral, (c) isosceles right angled, , + 4x2 + 4x + 1 = 0, -4x2 -4x-l = 0, - 4x2 - 4x -1 = 0, + 4x2 + 4x-l = 0, , (b) right angled, (d) obtuse angled, , 87. If sides of APQR are a, b sec C, c cosec C. Then, area of, i, , 2 71, 2 371, 2 571 ., sec — + sec — + sec — is, 7, 7, 7, (b)80, (c)24, (d) — 80, , 93. The value of, , APQPis, (b)2^(,’, , 4, , I, , 4, , For 0 < x <, , 7t, , (c)^?, (c)^c2, 4, , (a)-24, (d) -abc, 2, , Passage IV, , Passage II, , (Q. Nos. 94 to 96), , (Q. Nos. 88 to 90), , If 1 + 2sinx + 3sin2 x + 4 sin3 x+ ... upto infinite terms = 4 and, , let Pmn (x) = m log C05X (sinx)+nlogCOSJt (cotx);, , wherem,ne {1,2,...,9}, , - 3tt, number of solutions of the equation in ——, 4te is k., 94. The value of k is equal to, , [For example:, , (a) 4, , (b) 5, , P29 (x) = 2 log MJT (sin x) + 9 log cosjr (cot x) and, , P-n (*) = 7 log c«x (sin x) + 7 log, , 95. The value of, , (cot x)], , cos 2x -1, , On the basis of above information , answer the following, , (a)l, , (b)J3, , (c)2 -V3, , (d)3, , 88. Which of the following is always correct?, , (b)Pjx)^nVm>n, (d)2Pmn(x)<rnV m<n, , 89. The mean proportional of numbers P49 — | and P94, , 4J, , 96. Sum of all internal angles of a A:-sided regular, polygon is, , 71, 4, , (b) 4K, (d)2n, , (a) 5ti, (c)37t, , Passage V, , is equal to, , (b) 6, (d) 10, , (a) 4, (c) 9, , is equal to, , sin2x, , questions:, , (a), m*n, (c)2Pmn(x)<n V m<n, , (d)7, , (c) 6, , 90. If PM (x) = P22(x), then the value of sin x is expressed as, f t—, , , then (p + q) equals, , < p, (a) 3, (b) 4, (c)7, (d) 9, Note Mean proportional of a and b (a > 0, b > 0) is -Jab ], , (Q. Nos. 97 to 98), Let a is a root of the equation (2 sin x - cos x), (1 + cos x) = sin2 x, P is a root of the equation, , 3 cos2 x - 10 cos x + 3 = Oand y is a root of the equation, 71, 1 - sin 2x = cos x - sin x, 0 < a, 0, y < —., , 97. cos a + cos P + cos y can be equal to, (a), , Passage III, (Q. Nos. 91 to 93), If 70 = (2n + 1) n, where n = 0, 1,2, 3,4, 5, 6. then answer the, , (c), , 3-^6 + 2-^2 + 6, 6^2, 3-73 + 2, , 3^3 + 8, (b), , 6, , (d) None of these, , 6, , 98. sin(a - P) is equal to, , www.jeebooks.in, following questions., , 91. The equations whose roots are cos —, cos —, cos — is, 7, 7, 7, , (a)l, 1 -2^6, (c), 6, , (b)0, y/3-2y/2, (d), 6

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www.jeebooks.in, 96, , Textbook of Trigonometry, , g Trigonometric Functions and Identities Exercise 5:, Matching Type Questions, , 101. Match the statement of Column I with values of, Column II., , 99. Match the statement of Column I with values of, Column II., , (A) If 9 + 0 =—, where 9 and 0 are, , positive, then (sin 9 + sin 0) sin, , 1, , (P), , (<)“, , Column II, , Column I, , Column II, , Column I, , (A) The number of solutions of the equation, | cot x | = cot x + —— (0 < x < 7t) is, sin x, , (p) no, solution, , (8) If sin 9 + sin 0 =, , (q) 2, , and cos 9 + cos 0, , 3, , always less than, (q), , 2, , 9+ 0, is, = 2, then value of cot, 2, , (C) If 3 sin 9 + 5 cos 9 = 5, (9 * 0) then the (r), value of 5 sin 9-3 cos 9 is, , 3, , ( 71, (C) The value of sin2 a+ sin, ---- a, , (s), , 4, , (t), , 5, , (B) If sin 9 - sin 0 = a and cos 9 + cos 0, = b, then a2 + b2 cannot exceed, , 100. Match the statement of Column I with values of, Column II., , Column II, , Column I, , (A) If maximum and minimum values of, 7+ 6 tan 9- tan2 9 . „, ,, ---------------t-------- for all real values, , (p) X + pi = 2, , 1, , ., , (it, , 1., , U, , J, , (s), , 2, , (0, , 4, , sm — + a is, , (D) If tan 9 = 3 tan 0, then maximum value, of tan2 (9 - 0)is, , 102. Match the statement of Column I with values of, Column II., Column 11, , Column I, , (1+ tan2 9), , (r), , 13, , (A) In a t^ABC, sin (y j, , 71, , (P) - 1 + 4 sin n + A A sin, , 4 J', , of 9 - y are X and p respectively, then, , ( it + B cos C it +c, 7, I 4, , ■ M, + sin, — + sin M, — =, (q) X - p = 6, (B) If maximum and minimum values of, (, 71A, 5 cos 9+3 cos 9 + — + 3 for all real, , I, , 3j, , UJ, , (B) In a &ABC, sin ^yj, , values of 9 are X and p respectively,, then, , (C) If maximum and minimum values of, 71 - 91 for all, 1 + sin| — + 91 + 2 cosf —, 4, real values of 9 and X and p, respectively, then, , (q) 4 cos fit + A cos n + B, 4, A 4 ,, ti-CA, cos, , (r), , X+p=6, , (C) In a &ABC, cos ^y), + cos — - cos —, , 12J, , 12J, , (r) 1 + 4 sin it - A, 4, n-C, 7t - B, sin, sin, 4, 4, (s), , (s), , X—p = 10, , (t), , X-p = 14, , 4 J, , - 1 + 4 cos, , 71 -A ], , 4 J, , n-C, n-B, sin, cos, 4, 4, , (t), , 1 + 4 cos, cos, , 7t +A, 4, , it -C, it + B, sin, 4, 4, , www.jeebooks.in

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 97, , g Trigonometric Functions and Identities Exercise 6:, Single Integer Answer Type Questions, 1, , 103. In a AABC,, , 1, +—~"k, 1 + tan — 1 + tan2 2, 2, , 1, 4---------------2 B, , 114. If sum of the series 1 + x log., , I COSX I, , 1 + tan2 —, 2, A B, C*, 1 + sin — sin — sin — , then the value of k is, 2, 2, 2, , a -0, , a +fr, , sin, • cos, • cos 5, 2 >, cosy ,, \ 2 ,, 104. If^L:=---------- is, -, then —S + y'', sin P cos 8, f8-y, •sinP, •sin, sin, 2 7, I 2, equal to, c* ', , 371, , 71, , 571, , 771, , + x2 log, o ,1 - sin x;,, I, I, , 1 + sin x, I, cos x, , 1/4, , + ... o°, , fc(l -x), (wherever defined) is equal to ------- then k is equal, to, , 115. if, COS0, , +, , sin0, , = 56, , cos2 0, , 2, , 105. Find the exact value of the expression, , , . • \>/2, 1+sinx, cosx ., , r, , sin2 0, , = o then the, , 2, , value of (9x)3 + (5y)3, , is, , 971, 971, , tan-----tan — + tan-------tan — + tan —., 20, 20, 20, 20, 20, 44, , y, cos ri,o, 106. Letx = ^------ -, find the greatest integer that does not, ^sin n°, n=1, , exceed., , 116. The angle A of the AABC is obtuse., x = 2635 - tan B tan C, if [x] denotes the greatest integer, function, the value of [x] is, , 717. If4cos 36° + cot | 7, k 2 ., , =A +, , + y[n^, then the value of, , 107. Find© (in degree) satisfying the equation,, tanl5°- tan25°-tan35° = tan0, where 0g(O, 45°), , n2 + n, + n4 +, , n2 must be, i ■> i, , 118. Ifsin2 A =xand 11 sin(r A) = ax2 + bxz +cx* + dxi,, r■ 1, , 108. Find the exact value of cosec 10° + cosec50° - cosec70°., I, , TC i, , 109. If cos 5a = cos5 a, where a G 0, — , then find the, k 2/, possible values of (sec2 a + cosec2a + cot2 a)., 110. Compute the value of the expression, 2 7t, 2 2n t j 3n, r 2 7n, tan —+ tan— + tan— + ... + tan —., 16, 16, 16, 16, 111. Compute the square of the value of the expression, 4 + sec 20°, , cosec20°, cos B, 112. In AABC, if, , then the value of, 3, 3, 2, sin A, cosB, tanC, ------- + ------- +, is, cot 2A cot2B cot2C, 113. Let f and g be function defined by /(0) = cos2 0 and, , g(0) = tan2 0, suppose a and 0 satisfy 2/(a) - g(0) = 1,, then value of, - g(a) is, , then the value of 10a - 7b + 15c - 5d must be, 119. If x, ye fl satisfies (x + 5)2 +(y-12)z=(14)2 .then the, minimum value of -Jx2 + y2 is........, , 120. The least degree of a polynomial with integer coefficient, whose one of the roots may be cos 12° is, , 121. 1tA + B+C = 18®>,, , +, =*, sin A+sinB+sinC, , X, 2, , B, C, sin —sin — then the value of3fc3 + 2k2 + k +1 is equal to, 2 2, 122. The value of /(x) = x* + 4x3 + 2xz - 4x +7, when, lilt ., X = cot ----- IS.........., , 8, , 123. In any AABC, then minimum value of, v-,, J(sin A), 2020 V, =—y, ----- -■ — must be, (-/(sin B) + ^(sinC) - ^(sin A), , www.jeebooks.in, 124. If sin 0 + sin2 0 +sin3 0 = 1, then the value of, , cos6 0-4 cos4 0 +8 cos2 0 must be

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www.jeebooks.in, 98, , Textbook of Trigonometry, , 125.1, , I cos 0 - cos —, , I, , „, 3n, cos 0 - cos —, 8, , 8, , cos 0 - cos —, 8, , 77t I, , cos 0 - cos — = X cos 40, then the value of X is, , 126. If—1— + -—------ = 2k cos 40°, then, sin 20° V3 cos 20°, 18Zc4 + 162/c2 4- 369 is equal to, , 8 ), , Trigonometric Functions and Identities Exercise 7:, Subjective Type Questions, 127. Prove that tan 82-° =(-^3 + V2)(V2 + 1), 2, , or, , cot7-° =, 2, , + JI + V? + A, , 128. If msin(a + p) = cos(a - P), prove that, 2, 1, 1, -I------------------------------l-msin2a l-msin2P 1-m2, , 136. If{sin(a ~P) + cos(a-+ 2P)sinP}2 = 4cosasinPsin(a +0)., i, n, tan p, Then, prove that tan a + tan p = —- ------------- ;, , (V2cosP-l)2, , a. p 6 (0. 71 / 4)., , 137. If A, B, C are the angle of a triangle and, sin B, sin C, sin A, cos A, cos B, cosC - 0, then show that AABC is, cos3 A, , 129. Ifa+P + y = 7tand, , tan-(p + y - a)tan-(y + a ~P)tan-(a + p - y) = L, 4, 4, 4, then prove that 1 + cos a + cos p + cos y = 0., , cos3 B cos3 C, , an isosceles., 138. In any AABC, prove that, Vsin A, , 130. Find the value of a for which the equation, sin * x + cos4 x = a has real solutions., 131. If a and b are positive quantities and a > b, then find the, minimum positive values of asec0 — Btan0., , 132. If a, b. c and k are constant quantities and a, P, y are, variable subjects to the relation, a tan a + b tan p + c tan y = k, then find the minimum, value of tan2 a + tan2 p + tan2 y., , 133. if, , x, =, y, =, z, prove that:, tan(0 + a) tan(0 + p) tan(0 + y), X + ^sin2(a -p) = 0., x-y, , >3, , Vsin B + VsinC - Vsin A, and the equality holds if and only if triangle is, equilateral., , 139. If 2(cos(a - P) + cos(P - Y) + cos(y - a) + 3 = 0, prove, , ,, , c?P, , da, , that ----------------------- + ■------------------------ +, sin(P + 0) sin(y + 0) sin(a + 0) sin( y + 0), , dy, ------------ i----------- = o,, sin(a +0)sin(P +0), , where, ‘0’ is any real angle such that, a + 0, P + 0, Y + 0 are not the multiple of 7t., 140. If the quadratic equation, 1, , 134. Let at,a2,... ,an be real constants, x be a real variable, and /(x) = cosfflj + x) + - cos(a2 + x) + — cos(a3 + x) +, 2, 4, , - + i-C0SK+x>, Given that f(x}) = /(x2) = 0, prove that x2 - x, = nrn, for some integer m., 135. Eliminate 0 from the equations, tan(n0 + a) - tan(n0 + P) = x and, cot(n0 + a) - cot(n0 + P) = y., , 4«sect2‘a«x2 +2x +1 p2 - P + - | = 0have real roots, then, 2j, find all the possible values of cos a + cos'1 p., , 141. Four real constants a,b,A,B are given and, /(0) = 1 - qcos0 - bsin0 - A cos20 - Bsin20. Prove, that if y(0) > 0, V 0 e R, then a2 + b2 < 2 and A2 + B2 < 1., ..n ..cos 9, sin0. cos0n sin0n, ,, n, 1A, 142. If------ - +----- - =------ 2- +------2- = 1, where 0, and0o, cos02 sin02 cos02 sin02, do not differ by an even multiple of 7t, prove that, cos0,-cos0, n sin0.-sin0, n, 1, _ |, 1, V —, J, , www.jeebooks.in, cos2, , 02, , sin202

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, n, , 143. Prove that, , ^nCrarbn~r cos(r B -(n - r)A) = c"., , n —1, , ^nCk [cos k x.cos(n + k)x + sin(n - k)x.sin(2n - fc)x] =, 1=1, , r•0, , 147. Resolve zs +1 into linear and quadratic factors with real, (2" -2)cosnx., , 144. Determine all the values of x in the interval x E [0,2lt], , k, , = 3, where [.] represents the greatest, , 6, , ..f |, , 148. Prove that the roots of the equation, o 3 „ 2 , „, n, It, 371, 57t, ., 8x - 4x - 4x + 1 = 0are cos—, cos— ,cos — and, 7, 7, 7, , 145. Find all the solutions of this equation, x2 - 3 sin x----, , coefficients. Hence, or otherwise deduce that,, 71, 4 sin — -cos, = 1., 5, uoj, , 1, , which satisfy the inequality, 2 cos x < | -Jl + sin2x - ^/1 - sin 2x | < -Ji., , integer function., 145. In a AABC, prove that, , i, , 99, , hence, show that sec — + sec — + sec — = 4 and deduce, 7, 7, 7, 7t, 3n, 2 571, the equation whose roots are tan2, tan2 —, tan, 7, , g Trigonometric Functions and Identities Exercise 8:, Questions Asked in Previous 10 Years Exam, , I, 149. Let a and p be non-zero real numbers such that, , (a) 1 -, , 2(cos P - cos a) + cos a cos P = 1. Then which of the, following is/are true?, [More than one correct option 2017 Adv.], a, (a) 73 tan, - tan P = 0, 2, 2, (b) tan( — ] - 73 tan [ P] =0, ' ', <2j k2; <2j, , Ji, , (b) 1 +, , (c) 1-, , 2, , 4, , (d) 1 +, , 4, , 153. The number of all possible values of 0, where 0 < 0 < 7t,, for which the system of equations, (y + z)cos30 =(xyz)sin30, 2cos30 2sin30,, x sin 30 =----------+---------z, y, , and (xyz)sin30 = (y + 2z)cos30 +ysin30 have a, solution (x0, y0, z0) with yozo *0, is, [Integer Answer Type 2010], , (T, , (c) tan I — + 73 tan, =0, 2J, 2,, a, (d) 3/3 tan, + tan 'P = 0, 2, <2, , 71, 71, 150. Let — < 0 <----- . Suppose a j and P, are the roots of, 6, 12, the equation x2 - 2xsec0 + 1= 0, and a, andP, are the, , 71, 154. For 0 < 0 < —, the solution(s) of, 2, 6, , Y cosec, 4, , m =1, , a2 >p2, then a, +P2 equals to, [Single correct option 2016 Adv.], (a) 2(sec0- tan0), (b) 2sec 0, (c) -2 tan 0, (d)0, 13, ______ 1_, 151. The value of V —-—, is equal ’, kit, A sin fit-It , (k - l)n sin. (it- + —, 14, 6, 4, 6, , ., , [Single correct option 2016 Adv.], (b) 2(3-73) (c) 2(73-1) (d) 2(2+ 73), , 2, r, 152. Let f :(—1, 1) —> R be such that f (cos 40) = ------------ for, 2-sec2 0, , ), , cosec [ 0 + — I = 43/2 is/;are, k, 4 J, , [More than one correct option 2009], , roots of the equation x2 +2xtan0 -1=0. If oc 2 >Pj and, , to, (a)3-73, , '3, , «’ £, , (d> Tt, , sin4 x cos X, 155. If --------- 1= -, then, 2, 5, 3, Z 3, , 2, , 2, , (a) tan x = 3, 1, (c) tan2 x = 3, , [Single correct option 2009], sin*, x cos* x, 1, (b) ------ +-------- =—, 8, 27, 125, sin*, x, cos*, x, 2, (d) ------ +--------= —, 8, 27, 125, , ( 71, 155. Let 0 e 0, — and = (tan 0) tane,t2 = (tan0)co'9, k 4), ^=(0010)*“9 andt4 = (cot0)"“e, then, , www.jeebooks.in, 0e|O, —, I 4j, , ,, It , —, It I Then, the value(s) of f\/1\, - | is/are, l4, 4' 22), UJ, k3j, [More than one correct option 2012], b, , (a) t, > t2 > t, > t4, (c) t3 > t, > > t<, , [Single correct option 2006], (b) t4 > t, > t, > t2, (d)t2>tJ>t1 >t4

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www.jeebooks.in, 100, , Textbook of Trigonometry, , 157. The number of ordered pairs (a, P), where a, P G (-n, 7t), , 4, 5, 163. Let cos (a + P) = - and sin (a - P) = —, where, , satisfying cos (a - P) = 1 and cos (a + P) = 2 is, , 71, 0<a, P < —.Then, tan2a is equal to, , e, [Single correct option 2005], , (a) 0, (c)2, , . 56, (b) ~, 33, , . . 25, (a) —, 16, , (b) 1, (d)4, , , . 19, c —, 12, , [2010 AIEEE], (d)^, , 164. Let A and B denote the statements, II. JEE Mains and AIEEE, , A: cos a + cos P + cos y = 0, , 158. 5(tan2 x - cos2 x) = 2 cos 2x + 9, then the value of, , cos 4x is, 3, (a)-i, , B: sin a + sin P + sin Y = 0, , [2017 JEE Main], 7, (d)--, , 12, (b)l, (0(b) 7, , 3 If cos (P - y) + cos (y - a) + cos (a - 3) = —, then, 2, [2009 AIEEE], (a) A is true and B is false, , 159. If fk (x) = 2 (sin* x + cos* x), where x G R, k > 1, then, ”, , (b) A is false and B is true, (c) Both A and B are true, (d) Both A and B are false, , k, , [2014 JEE Main], , (x) is equal to, , f<, , (b) 1, , (a) 1, 6, , (c) 1, 3, , 4, , tan A, cot A, 160. The expression --------------4---------------- can be written as, 1 - cot A, 1 - tan A, [2013JEE Main], (a) sin A cos A + 1, (c) tan A + cot A, , 165. A triangular park is enclosed on two sides by a fence, and on the third side by a straight river bank. The two, sides having fence are of same length x. The maximum, area enclosed by the park is, [2006 AIEEE], Jy, (b) | x2, (c) nx1, (d) | X2, (c) nx2, (a), (b) 7, , (b) sec A cosec A + 1, (d) sec A + cosec A, , 161. In a APQR, if 3 sin P + 4 cos Q = 6 and, , 4 sin Q + 3 cos P = 1, then the angle R is equal to, [2012 AIEEE], (a), , (b) ~, 6, 371, (d) —, 4, , 6, ,. n, (c) 4, , (b) 1 < A < 2, , P, , 71, 167. In a APQP, Z R = -. If tan, 2, , 162. If A = sin2 x + cos'* x, then for all real x, (a) — < A < 1, 16, 3, 13, (c) - < A < —, 4, 16, , 166. If 0 < x < n and cos x + sin x = -, then tan x is, AIEEE], 2, [[2006, ’, (4, +, V7), (4-^7), (b) (a), 3, 3, (1~V7), (1 + ^7), (d), (c), 4, 4, , and tan, , Q, , of ax2 + bx + c = 0, a * 0, then, , [2011 AIEEE], , 3, (d) - < A < 1, 4, , are the roots, , 2, , 2, , [2005 AIEEE], , (a) b = a + c, , (b) b = c, , (c) c = a + b, , (d) a = b + c, , Answers, Exercise for Session 1, 1. 72°, 18°, , 2. — cm, 2, , ‘•&I, , 8.880cm/s, , 7. 252 cm, , 3. 70 m, , 4. 45°, , 5. 8k, , 4. ± y]a2 + b2 - c2, , a2, , io.2, , 9. 10, , 5, , b, , Exercise for Session 4, 9. 1.7 cm, , 10.7, , 1.28, , 4-, , Exercise for Session 2, 3.-3, , ’•4+4=2, , 5. 13, , 6.2, , 7.—, 12, , 3.-2, , 1., 0, , y= sin|, -i—, , Jt/2, , 3n/2, , y = sinx, z27t, , 5n/2, , 3?e, , 10.4, , www.jeebooks.in, Exercise for Session 3, 1., , 2k, , k2 + \, , 2. 1, , 4.x=l,y=0, , 6.0

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 5., , Exercise for Session 11, , y, , ■1 ■, , 0, -1* •, , n/2, , n, , 3n/2 2n, , 5nJ2, , 8. - V2 and 41, , From the graph, the period of the function is n., , i.(b), 7.(b), 13. (d), 19. (a), 25. (d), , B’T, y4, y = sin x, , n, , 3n, 2, , 2n, , 0, , *x, , 2, , y = cos x, , y, , 9., 3, 2, 1, ■nJ3 -nJ6, , 0 n/6, , — 1—, , 7nJ&, , nJ2 '2nJ3, , •X, , ■2, , -3, , 10.1, 4, , Exercise for Session 5, 1.V3, 7.2, 10.0, , 2. Negative 3.-1, , 4. 1, , 5. 1, , 6. 1, , 9.2, , 8. does not exists any real solutions, , 6. a1 + b1, , 3.—, 4, 8. 27, , Exercise for Session 7, 4.1, 5.0, 6. 2 cot", 5, , Exercise for Session 8, 3.-3=, 1. False 2. tan 0, Vs, 7.-1, , 8. 1, , 9. 15, , Exercise for Session 9, 3+2V2, 4, l.-2± V52., 8, 43, Exercise for Session 10, 4.1, 5.2, 7.-1, , 4. (a), 3. (a), 2.(b), 5. (b), 8. (a), 10. (b), 11. (d), 9. (a), 14. (b), 16. (b), 17. (b), 15. (a), 20. (a), 22. (a), 23. (a), 21- (c), 26. (c), 28. (b), 27. (b), 29. (c), 32. (a), 35. (b), 33.(b), 34. (c), 31-(c), 40. (c), 38. (d), 39. (d), 37. (a), 41. (c), 44. (c), 46. (b), 43. (b), 45. (a), 42. (b), 50. (a), 51. (a), 52. (d), 49. (a), 48. (b), 55. (a,b,c) 56. (b,d) 57. (b,c), 54. (a,b,c,d), 59. (b.c) 60. (a,b,d) 6l.(b,c) 62. (c,d) 63. (d), 67. (a, b, c, d), 65. (b, c) 66. (a, b, c, d), 70. (b, c) 71. (a, b, c, d), 69. (a, b, c, d), 76. (b), 74. (b, c) 75. (a), 73. (a, b, c, d), 82. (c), 81. (a), 78. (d), 80. (a), 79. (d), 87. (a), 88. (b), 86. (b), 85. (b), 84. (b), 94. (b), 93. (c), 90. (c), 92. (a), 91. (b), 98. (c), 96. (c), 97. (b), 99. A—(p, q, r, s, t); B—(s, t); C—(r), 100. A—(r, s); B—(r, t); C—(p, q), 101. A—(r); B—(p); C—(p); D—(q), 102. A—(r, t); B—(p, s); C—q, 103. (2), , 104.(1), , 109.(5), , 110.(35), , 4.H, , 7.1, , 112. (2), , y, , 8, , 6.31, 7, , 10.1, , n, 10. cot—, 5, , 107. (5), , 108. (6), , 113.(1), , 114.(2), , 117.(91) 118.(3448), , 119.(1), , 120.(4), , 121.(1673), , 122.(6), , 124. (4), , 125. (2), , 126.(1745), , 140. cosct-cos 'P =, , 123.(6060), , 131. Minimum value is 4a1 - b1, , 132. Minimum value is, , 8.3^, , 8, , 5. a = - I, P = 3, , 4. 2 cos nQ,, , 111.(3), , 47. (c), 53. (a), 58. (b,c), 64. (b, c, d), 68. (c, d), 72. (b, c), 77. (b), 83. (b), 89. (b), 95. (d), , 116. (2634), , 135. cot(a+p) = l, x, 2, , 106. (2), , /■, , 5. tan p + 2 tan, , 33, , A-B, , 105. (5), , 6. (a), 12. (d), 18. (b), 24. (b), 30. (b), 36. (b), , 115. (3136), , 130.3 < a<, 1, 2, , Exercise for Session 6, 1.1 and IV, , 10.13:4 or 1:4, , Chapter Exercises, , 7 n 5n’, 8., , ' positive, ......... 6., . -3 < A £ 1, 5., , 4.2, , 2.0, —3n/2 -n -n/2, , 101, , A2, , ], , ly + z>2 + <, , 1, , y, , — 1, when n is an even integer, 3, — + 1, when n is an even integer, 3, , n In, —, .4 4 ., 145. Only two solutions, x = 0,43, 144. xe, , 148. Required equation is, zJ - 21z2 + 35z- 7 = 0 whose roots are, 2n, 2 3n , :• 5n, tan —, tan —, tan —, 7, 7"1, 150. (c), 151. (c) 152. (a,b) 153.(3), 149. (b, c), 157. (d) 158. (d) 159. (d), 156. (b), 154. (c,d) 155. (b), 162.(d), 163. (b) 164. (c) 165. (b), 160. (b) 161. (b), , www.jeebooks.in, 2, , 8. 1, , 166. (b), , 167. (c)

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , => 2 cos, , 3rc, 2, , 3C, 3n, ■ 2 cos, 2, 2, , 3B, 2, , ■ cos, , 3%, 2, , 3A, =0, 2, , 105, , =>, -5 <3 cos2x + 4sin2x <5, .•.Minimum of 3 cos 2x + 4 sin 2x = - 5, So,, , I, I, |, -, , min f(x) = 3“5 =—, 243, 32. Let AB = x and ZBDC = a, , =>, , q_, , 3A, 3B, 3C, — = n or — = 7t or — =7t, 2, 2, 2, 2tc „ 2tc, 2n, A = — or B = — or C = —, 3, 3, 3, , ■, , ■, , C, , D, a, 0, , JpP+q2, , P, , 29. •: | tan A | < 1, , -1 < tan A < 1 and 0 < | A | < y, , 7(i+s^n 2^)=, , A, , X, , In ADAM, tan(7t - 0 - a) =, , II + tan A | _ (1 + tanA), , 7(1 + tan2 A) 7(! +, 7 2 tanA7 \, , =>, , J + tan2 A;, , tan(0 + a) = —- - => q - x- p cos(0 + a), 9-x, , 9, , 7(1 + sin 2A) + 7(1 ~ sin 2A) _, 2, -------- = cot A, 7(1 + sin 2A) - 7(1 ~ sin 2A) 2 tan A, , q, , cos0, , p, , sin 0, , q cos©', ~P, , ?+, P, , 30. Let /(0) = cos2(cos 0) + sin2(sin0), , yp sin0,, cos, c_° 0, sin 0, , +P, , q, , v cot a = —, Pj, , COS0, , p, , -1 < cos 0 < 1 and -1 < sin 0 < 1, cos 1 < cos(cos 0) < 1 and - sinl < sin(sin 0) < sin 1, cos21 £ cos2(cos 0) S 1 and 0 < sin2(sin 0) < sin21, Maximum value of /(0) = 1 + sin2 1, , 3, 4, Let — = sin <|) and - = cos 0, = 3^(sin^ *, , For minimum value of given function, sin(0 + 2x) will be, minimum., i.e., sin(0 + 2x) =-1, , = cot a • tan a = 1, Similarly, cot 2a cos(2n - 2)a = 1,, cot 3a • cot(2n - 3)a = 1,.... cot(n -1 )a cot(n + l)a = 1, Thus cot a cot 2a cot 3a ... cot(2n - l)a, = {cot a cot(2n - l)a} {cot 2a cot(2n - 2)a}, ... {cot(n - l)a cot(n + l)a} • cot na, = 1 • 1 • 1... 1 • 1, , /(x)=35(“n= —, , 243, Alternate Method, /(x) = 27c0‘2x 81”n 2x = 34, , (g2 + p2)sin0, q sin 0 + p sin 0, , Now cot a • cot(2n - l)a = cot a cot I — - a, \2, , 4, 5, , 5 I - coi 2x + - sin 2x, , /(x) = 35<sin * cos 2x + cos * “, , sin0, sin00, sin, q sin 0 + p cos 0, psinO, , 33. Given 4na = ft => 2na = —, 2, , 31. Let /(x) = 27CO5 2x 81"n 2x = 33co’2x+ 4 $in2x, , Thus, , cot 0 cot a -1, cot a + cot 0, , _ g (cot a + cot 0) + p (cot 0 cot a) + p, cot a + cot0, , (1 - tan A), 7(1 - tan2 A), , =3, , x-q, , => x = q- p cot(0 + a) = q - p, , 11 - tan A |, 7(1 + tan2 A), , 3, S, , B, , q, , M, , 2 tan A, I +--------5—, 1 + tan A, , and 7(1 - sin 2 A) = 1 -, , p, , [, , n-(0+a), , n t n, --<A <2, 2, , I, , 7t, , v cotna = cot — = 1, 4, , =1, , 34. We have (sin .A +sin B + sinC)(sin A +sin B - sinC), , www.jeebooks.in, 2x = 33, , 2x + 4 nn 2x, , For minimum value of given function, 3 cos 2x + 4 sin 2x will, be minimum., :. - 732 + 42: 3 cos 2x + 4 sin 2x < Jl2 + 4'2, , =>, , = 3 sin A sin B, (sin A + sin B)2 - sin2 C = 3 sin A sin B, sin2 A + sin2 B -sin2 C =sin A sin B

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www.jeebooks.in, 106, , Textbook of Trigonometry, , =>, , sin" A + sin(B +C) sin(B -C) =sin AsinB, , sin A [sin(B + C) + sin(B -C)] = sin A sin B, [■.■ A + B + C = jt], sin A(2 sin B cos C) =sin A sin B, , cos C - - => C = 60°, 2, 35. From the third relation we get, cos 0 cos 0 + sin 0 sin 0 = sin P siny, => sin2 0 sin2 0 = (cos 0 cos 0 - sin P sin y)2, , \, , sin2 p, , = - - (1 - cos 2) = - - (2 sin21) = - sin21, 2, 2, This shows that y = - sin21 is a straight line which is parallel, (K, , 39. f(d) = | sin 0 | + | cos 0 |, V 0 e R Clearly, /(0) > 0., , Also,, , _f2(0) = sin2 0 + cos2 0 + | 2 sin 0 • cos 0 |, = 1 + | sin 20 |, 0 < | sin 20 | £ 1, 1 </2(0)<2=>l </(0)<a/2, , 1, , 40. A = cos(cos x) + sin(cos x), , = sin2 P sin2 y (1 - sin2 a)'.2, , = -72 < cos(cos x) cos — + sin(cos x) sin — f, I, 4, 4 I, , => sin4 a(l - sin2 p sin2 y), , K, cos X---4., , - sin2 a(sin2 P + sin2 y - 2 sin2 p sin2 y) = 0, . 2, , sin a, , and cos2 a =, , sin2 P - sin2 y - 2 sin2 P sin2 y, , 1 - sin2 psin2 y, , [•.• sin a * 0], , -1 < cos cos x - — I < 1, I, 4?, -■j2<A<4i, , 1 - sin2 P - sin2 y + sin2 p sin2 y, 1 - sin2 Psin2 y, sin2 p - sin2 p - sin2 y + sin2 y - sin2 P sin2 y, , tan2 a =, , cos2 p - sin2 y (1 - sin2 P), sin2 P cos2 y + cos2 P sin2 y, , 41. We have, — = tan nd, 'Vn, an(j Vn ~ V”"1 - cosn 9 sec" Q ~ cos(n - 1)9 secn~'0, sin(n - 1)0 secn-10, , cos2 P cos2 y, , cos n0 sec 0 - cos(n -1)0, sin(n -1)0, , = tan2 p + tan2 y, , cos nd - cos(n - 1)0 cos 0, cos0 sin(n -1)0, , => tan2 a - tan2 p - tan2 y = 0, , 36. tan P =, , n tana, , cos(n -1)0 cos 0 - sin(n - 1)0 sin 0, - cos(n - 1)0 cos 0, , 2, , 2, , 1 + tan a - n tan a, n tan a, 1 + (1 - n) tan2 a, , cos 0 sin(n - 1)0, , n tan a, 1 + (1 - n) tan2 a, tan a ■ n tan a, 1 + (1 - n) tan2 a, , tan a -, , => tan(a - p) =, , tan a + (1 - n) tan3 a - n tan a, 1 + (1 - n) tan2 a + n tan2 a, (1 - n) tan q(l + tan2 q), , 1 + tan2 a •, 97, , 2, , to X-axis and clearly passes through the point I —, -sin 11., , V, sin2 y _ sin P sin y, -sinpsin y, sin2 a, sin2 a J k sin2 a, >, [from the first and second relations], => (sin2 a - sin2 P) (sin2 a - sin2 y), , =>, , = ~[2 cos2(x + 1) - 1 + cos 2] - cos2(x + 1), , = (1 - n) tan a, , = - tan 0, v„-vn_, i[/n, tan nd, So, that ------------- F------ = - tan 0 +, *0, n, ^-1--- n Vn, 42. If a, b > 0, , Using A.M. > G.M., we get, - + ->-|=, a b Jab, =>, , 2, , /(x)>, , 7C, n, 1, cos ---- x < cos — + X, , (cos0) (sin0), = k, so that cos 0 = ak and sin 0 = bk. Then, a, b, a cos 20 + b sin 20 = a(l - 2 sin2 0) + 2b sin 0 cos 0, , = a - 2ab2k2 + 2b ■ bk ■ ak, , = a - 2ab2k2 + 2ab2k2 = a, , 6, , ), , .6, , 2____, , <, , 2 rc, .2, cos---- sm x, 6, 2, , 3, 4, , 2______, 1 - cos 2x, 2, , www.jeebooks.in, 38. Let y = cos x.cos(x + 2) - cos2(x + 1), 1, ,, = - [cos(2x + 2) + cos 2] - cos (x + 1), 2, , 11 cos 2x, - +------4, 2

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , 7C, , —1, , 47. Given that, a + p + y = n, , Now for 0 < x < —, — < cos 2x < 1, 3 2, cos 2x, =>, 2, 2, V4, 4, =>, , Taking a=-^;0 = -^- and y= 2n, , o< ET, , 43., , sin a + sin 0 + sin y = - 1 -1 + 0 = - 2, but sin a + sin 0 + sin y £ - 3 for any a, 0, y, Hence, minimum value of sin a + sin P + sin y is negative., 48. cos x - sin a cot 0 sin x = cos a, , 4 °° 1 ., LV3 J, , Since .f is continuous range of<'f' is, , =>, , 0 < sin2 0 < 1 and 0 S cos2 0 < 1, =>, , 0 < sin8 0 < sin2 0 and 0 < cos14 0 2 cos2 0, , :. 3 < sin4, , sinp cos x-sina cos0sin x=cosasin0, , => sin 0 11 - tan2 — | - sin a cos 0 ■ 2 tan —, 2, 2, , 0 < sin8 0 + cos14 0 < sin2 0 + cos2 0, Hence,, 0<A<l, AA . (3n, |, 44. sin----- a | = - cos a, sin, .2, k2, J, , = cos a sin 0, , + tan2, , j, „ „, , ), , 107, , X, , => tan2 - (- sin 0 - cos a sin 0) - sin a cos 0 • 2 tan —, 2, 2, + sin 0(1 - cos a) = 0, , = cos a, , sin(3n + a)=-sina, sin(5rc - a) = - sin a, y- -a^ + sin4 (3 k + a), , 4 [sin2 a cos2 0, , - 2 sin a cos 0 ± + sin2 0(1 + cos a), =>, , - 2 sin6^~ + a j + sin6(5rc - a), , |(1 - cos a)], 2 sin 0 (1 + cos a), , x, tan — =, 2, , -sina cos0, , ± -Jsin2 a(sin 2 0 + cos2 0), , = 3{cos4 a +sin4 a} - 2{cos6 a + sin6 a}, , sin 0(1 + cos a), , = 3{1 - 2sin2 a cos2 a} - 2{1 -3 sin2 a cos2 a} = 1, , 0, , (, it k, = -^2 sin x+ - + —, k, 6, 4,, , 5rc, , = -Ji sin, ., , 12., , => cot X, , | cot x cot y + 1, k cot y - cot x, , a, , a, , 0, , = tan - tan — or - tan — cot 2, 2, 2, 2, , 49. ■: cos4 0 sec4 a , - and sin4 0cosec2 a are in AP, 2, 1 = cos4 0 sec2 a + sin4 0 cosec2 a, , <45, , Equality holds when x + — = — ie, x = —, 12 2, 12, Therefore, maximum value of given expression is attained at, n, x=—, 12, 46. cot2 x = cot(x - y) • cot(x - z), 9, , sin a(l - cos 0 ± 1), sin 0(1 + cos a), , - sin a cos 0 ± sin a, sin 0(1 + cos a), , 45. sin| x+ — + cos l X + nk, 6., k 6., , cos4 0 sin4 0, 1 = —,2— +, cos a sin", sin a, , cos4 0 sin4 0, (sin2 0 + cos2 0)2 = —2~ +, cos a sin2 a, /, _ 1_, cos4 0, -1 + sin4 0, [sin2 a -1, xcos2 a, , n, , cot x • cot z + 1, cot z - cot X, , - 2 sin2 0 cos2 0 = 0, , =>cot 2 x-cot y • cot z - cot 3 x-coty-cot.33 xcotz + cot 4 x, , sin4 a cos4 0 + sin4 0 cos4 a, , 2, , = cot x • cot y ■ cot z + cot x • cot y + cot x ■ cot z + 1, , => cot x cot y(l + cot2 x) + cot x cot z(l + cot2 x), + 1 - cot4 x = 0, , - 2 sin2 0 cos2 0 sin2 a cos2 a = 0, =>, , (sin2 a cos2 0 - cos2 a sin2 0)2 = 0, , =>, , tan2 0 = tan2 a, , => cot x(cot y + cot z) (1 + cot2 x), + (1 - cot2 x) (1 + cot2 x) = 0, , => cot x(cot y + cot z) + (1 - cot2 x) = 0, , Now,, , and, , 0 = nn ± a, n e 1, cos8 0 sec6 a = cos8 a sec6 a = cos2 a, sin8 0 cosec6 a = sin8 a • cosec6 a = sin2 a, , www.jeebooks.in, cot2 x-1 1 ,, ., ,, = - (cot y + cot z), 2 cot x, 2, , =>, , 1 ., x, - (cot y + cot z) = cot 2x, , Hence, cos8 0 sec6 a, -, sin8 0 cosec6 a, 2, , ie, cos2 a, -, sin2 a are in AP., 2

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, 8a + 1, „ 25, 6 0,—, 16, 16, , = 1 + sin2x, , :., , P(18°) = P(72°) = (1 + sin218°) + (1 + sin272°), , I1, , 8a + 1 e [0,25], , = 1+ 1 + (sin218°+ cos218°) =3, , I, , Ko, 3 sin(a + P) - 4 cos(a + P), 3Jr. b =------------ y=----------------V3 sina, 3(sina cosp + cosasinP) - 4(cosa cosP + sina sinP), -Ji sina, , 109, , a e —,3, .8, , =>, , 62. A = sin 44° + cos 44°, = cos 46° + sin 46° = C, B = sin 45° + cos45° = -V2[sin90°], , = -4= for 0 < p < —, -Ji, 2, , A=Ji -4= sin 44° + -4= cos 44°, , V2, , LV2, , for — < p < it., 2, , 15, , J, , = 5/2 [sin44°- cos45° + cos44°- sin 45°], , 60. cot A = 4, , = -Ji sin89°, , -3, cot C = —, 4, , => B>A, 63. tan(2a + P) = x, tan(a + 2P) = y, , 4, , =>, , tan(3(a + P)). tan(a - P), = tan [(2a + p) + (a + 2P)]., , 3, , tan[2(a + P)-(a + 2P)], , => C is obtuse angle., 4, 3, sinC =-, cosC = 5, 5, , tan(2a + P) + tan(a + 2p), 1 - tan (2a + p). tan(a + 2P), , tan(2a + P) - tan(a + 2P), 1 + tan(2a + P). tan(a + 2P), , 13, 12, , x + y x-y, 1 - xy 1 + xy, , B, 5, , -12, tanBr, =---5, r, I2, =>, tanD = —, 5, => D is an acute angle, 12, 5, sinD = — cosD =—, 13, 13, Hence, sin(C + jD) =sinCcosD + cosC-sinD, , 64. We have x =, , 1 - sin 0, cos 0, , Multiplying, we get xy =, , x2 - y 2, x2y2, 1 - xy, , 1 + COS 0, sin 0, , (1 - sin (1 + cos 0), cos Q sin 0, , 1 - sin 0 + cos 0 - sin 0 cos 0, =>, , + sin 0 cos 0, cos 0 sin 0, , 1 - sin 0+ cos 0, cos 0 sin 0, , 20-36, 65, , -16, ~ 65, , Also, sin(A + B) = sin(2rc - (C + D)), = -sin(C+ D) = —., 65, , 61. 2| cos2x + -cosx = a, , I, , 2, , 2| cosx + -, , I, , |2, , 4. 1, , 1, , =a + -, , 8, , 1, 1 2 a, cosx + - = - + —, 4., 2 16, , and, , (1 - sin 0) sin 0 - cos 0(1 + cos 0), x -y —---------------------------------------cos 0 sin 0, sin 0 - sin2 0 - cos 0 - cos2 0, cos 0 sin 0, , sin 0 - cos 0 -1, cos 0 sin 0, , -(xy + 1), , Thus, xy + x- y + l = 0., ., 1+x, y-1, x =----- - and y =------ ., 1 —x, , 65. The given relation can be written as, • 2, 1 - sin x cos2 X, X, tan, sin x, sin x, 2, , www.jeebooks.in, 25, 1 2, cosx + - 6 0,—, 4, 16

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www.jeebooks.in, 110, , Textbook of Trigonometry, , => 2 sin2, , 2, , (i)., , 67. v 3 sin 0 = sin(2a + 0), , 2, , 2 sin 0 = sin(2a + 0) - sin 0, = 2 cos(a + 0) sin a, sin 0 = cos(a + 0) sin a, •••(*), Alternate (b) is correct, Also, sin0 = sin(a + 0)-a), = sin(a + 0) cos a - cos(a + 0) sin a, -(ii), From Eqs. (i) and (ii), sin 0 = sin(a + 0) cos a - sin 0, 2 sin 0 = sin(a + 0) cos a, (.’. Alternate (c) is correct), Alternate (a), LHS = (cot a + cos(a + 0))(cot 0-3 cot(2a + 0)), =>, , => 2 tan2, , 2y(l + y) = (1 - y)2, , =>, , where y = tan 2, , X, , 2, , y2 + 4y - 1 = 0, , 2, Since y > 0, we get, y = Vf5-2, , _ (V5 - 2)2 2 + %/5, V5 + 2 2+V5, , -f sin(2a + 0), , = (9-4V5)(2 + V5), J(cos 2A - sin 2A)2 + 1, , 66. y, , 3 cos(2a + 0) |, sin(2a + 0) ], , 3 sin 0, , cos 0, , 3 cos(2a + 0)', , (, , -J(cos 2A + sin 2A)2 - 1, ¥, , l^sin a • sin(a + 0), , cos 0, sin 0, , sin a • sin(a + 0) J sin 0, , ± (cos 2A - sin 2A) + 1, ± (cos 2A + sin 2A) - 1, , 3 sin 0, sin a • sin(a + 0), , 2 cos A-2sinAcosA, - 2 sin2 A + 2 sin A cos A, , cos A(cos A - sin A), sin A(cos A - sin A), , - (cos 2 A - sin 2A) + 1, -(cos 2 A + sin2A)-l, (1 - cos 2A) + sin 2A, - (1 + cos 2A) - sin 2A, 2 sin2 A + 2 sin A cos A, =-------- ----------------------- = - tan A, - 2 cos A - 2 sin A cos A, (cos2A-sin2A) + 1, Vi, - (cos 2A + sin 2A) - 1, (1 + cos 2A) - sin 2A, - (1 + cos 2A) - sin 2A, , 2 cos2 A - 2 sin A cos A, , - (cos 2A - sin 2A) + 1, (cos 2A + sin 2A) -1, , J, , 2 sin(a + 0) sin a, , sin 0, , Alternate (d), tan(a + 0) = 2 tan a, sin(a + 0) 2 sin a, cos(a + 0) cos a, , =>, , sin(a + 0) cos a = 2 cos(a + 0) sina, , =>, , => sin(a + 0) cos a - cos(a + 0) sin a = cos(a + 0) sin a, sin 0 = cos(a + 0) sin a, [Alternate (b)], , 68. Pn(u) be a polynomial in u of degree n., sin 2nx = 2 sin nx cos nx, = sin x P^ /cos x) or cos x P^ j(sinx), sin a -cos a, 69. tan 0 =, sin a + cos a, , .’., , tan a -1, tan a + 1, , cos A - sin A, cos A + sin A, , y< =, , sin 0, , =6, , - 2 cos2 A - 2 sin A cos A, , 1 - tan A, IIt, A, (it, 7C, = - tan I---- A I = - cot I — + A, 1 + tan A, \4, 4, , ,, , (v 3 sin 0 = sin(2a + 0)), cos 0 - cos(2a + 0)^, , 3 sin 0, sin a • sin(a + 0), , which gives us four values of y, say y,, y2 , y3 and y4 We have,, cos 2A - sin 2A +.1 (1 + cos 2A) - sin 2A, cos 2A + sin 2A - 1 (cos 2A - 1)+sin 2A, , 3 sin 0, , I, , K i, , tan 0 = tan a---I, 4J, , =>, or, , (1 - cos2A) + sin 2A, - (1 - cos 2A) + sin 2A, , n, , it, , 0 = htc + a —,ne I, 4, , 20 = 2nit + 2a - —, 2, , sin 20 = sin I 2a - — I = - cos 2a, , I, , 2J, , I, , 7t 1, , www.jeebooks.in, 2 sin2 A - 2 sin A cos A, , - 2 sin A + 2 sin A cos A, , and, , cos 20 = cos 2a---- = sin 2a, k, 2J

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www.jeebooks.in, 112, , Textbook of Trigonometry, , and sin 2A + sin 2B - sin 2C, - 2 sin(A + B) cos(A - B) - 2 sin Ceos C, = 2 sin C{cos(A - B) - cos C}, = 2 sin C{cos(A -C) + cos(A + B)}, = 2 sin C(2 cos A cos B), = 4 cos A cos B sinC, 77. v a cos x + b sin x = c, , 1 - tan2, , Adding Eqs. (i) and (ii), then, +~, a6 + a5 + a3 = -1, 2C = a + a2 + a4 ■*-, , (■/ sum of 7, 7th roots of unity is zero), , c = —-, , 2, Also, multiplying Eqs. (i) and (ii), then C2 +S2 = 2, , (’.• a7 = 1 and sum of 7, 7th roots of unity), , 2 tan, , a-, , -c, , + b{, , 1, 2, , S2 = 2-, , =>, , 2, , 7, 4, , 1 + tan2, , 1 + tan2, , 2, , (a + c) tan2, , + (c - a) = 0, , 80. We observe that y =81 sin2, , gpm2 x, , 2b, (a + c), , and, , Now,, , 8l2sin2x, , c-a, , (gisin2 x, , a+c, , + 81l-sin2x_30=?0, , -30.81, , -3) (81, , sin2 x, , sin2 x, , + 81 = 0, , -27) = 0, , , 1, . V3, sin x = ± - or sin x = ± —, 2, 2, , n , 5tc, ,71,271, x = ± —, ± —, or x = ± —, ± —, , 2b, a+c, = — = independent of c, c-a - a, 1a + c), , 6, , 3, , 3, , 6, , sin2 x, , cos2 X, , => The graph y =81sm x + 81c<” x-30, Intersects the X-axis at eight points in (- 7t < x S 7t)., => Statement-1 is true., 81. Statement-2 is correct, using it we have cos 3x = sin 2x, , cos x + b sin x < -/(a 2 + b2), b2) < c < ^(a2 + b2), , 4 cos3 x - 3 cos x = 2 sin x cos x, , =>, , 78. v A + B + C = 180°, , Similarly 4 cos3 y - 3 cos y = 2 siny cos y, , A = 180°-(B + C), tan A = tan(180° -(B + C)), , tan B + tanC, = - tan(B +C) = 1- tan B tanC, , So,, , 4(1 - sin2 x) - 3 = 2sin x, , =>, , 4sin2 x + 2sinx-l=0, , and, , 4 sin2 y + 2siny-l = 0, , Hence, sin x = sin 18° and sin y = sin(- 54°) = - sin 54° are the, roots of a quadratic equation with integer coefficients., 82. The minimum value of the sum can be - 3 provided, sin a = sin P = sin y = -1, , tan B + tan C, tan B tan C - 1, Now, v A is obtuse, tan A < 0,, then, tan B + tan C >0, tan B tan C - 1 < 0, =>, tan B tan C < 1, 79. Let 5 = sin (y J + sin (y) + sin (y j, , and, , -30 = 0, , + 81cos, , sin2 x = - or 4 4, , tan, , Also,, , ,2, , x, , a - (41 -1) ~ p = (4m -1) 5 1 =(4n -1)5, Li, , It, , a + p + y = tc => [4 (/ + m + n) - 3] — = K, 2, => 4(/ + m + n) = 5 which is not possible as I, m, n are integers., 1. minimum value can not be - 3., , Now, , „, , C = COS (y) + COS (yj + COS (yj, C + iS=a + a2 + a4, , (i), un,, f271^1 . . <27^, Where a = cos I “I +1 sin I ~\ is complex 7th root of unity., , „, , 3tc o, , 3tc, , n, , n, , But for a = —, p = —, y = - 2k, a + P + y = tc, 2, 2, and, sin a + sin P + siny = 2, So, sin a + sin P + sin y can have negative values and thus the, minimum value of the sum is negative proving that, statement-1 is correct. But the statement-2 is false as, , www.jeebooks.in, Then,, , 3tc, , C-iS = a + a2 + a4, , = a6 + a5 + a 3, , ...(ii), , a + P + y = nfora=P = —, y = - 2n which are not the, 2, angles of a triangle.

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , ( 0', 83. We have 2 sin \,2., cos, cos, , 'o, , 85. vA+B=3, , +sin, , 2, , 0, , • f01 + I cos T0—1, , 0, , + sm -, , U;, , |, , I2J, , 2, , -sin, , tan A tanB = 1 - -^= (tan A + tan B), , /.tan A tan B will be maximum if tan A + tan B is minimum., But the minimum value of tan A + tan B is obtained when, tan A = tan B, , 7t i. -0 n - n ,, i, (e-'=>, sin - + — > 0 and cos - + — < 0, 2’, 4, U 4J, re 0 n „, =>, 2nn + -<- + — > 2nn + re, 2 2 4, „, 71 0 „, 3n, 2nn + — < - < 2nn + —, 4 2, 4, So statement-1 is true but does not follow from statement-2, which is also true., 84. 2 cos 0 + sin 0 = 1, , 2 1 - tan2 Ik2, , 1 + tan2, 3 tan2, , 0, 2, , 2J, , 2, , =>, , -1 = 0, , b, , 0, , 1 A n, = - - as 0 * —, 2, 3, 2, , /A, , From figure, it is clear that a = b secC = c cosec C, , Equilateral triangle => Area =, , -1, , 0, , -"l, + 12 tan, 2j, , 1 + tan2, , 1, 7-7X- + 12, 9, , 2, , 1+1, , -i, 2J, , 9, , Showing that statement-1 is true., In statement-2, cos 20 - sin 0 = 2, 2(1 - 2 sin2 0) - sin 0 = 1, , =>, , sin 9, , -2 ±^4+ 16, , - 1 ± V5, , 8, , 4, , 75-1, , ,, , 0 = 18°, 4, cos 60 = cos 108° = cos(90° + 18°), = -sinl8°, sin 0 + cos 60 = 0, So statement-2 is also true but does not lead to statement-1., sin0 =, , 2, , 4, So/. (Q. Nos. 88 to 90), 70 = (2n + l)rt,n = 0to6, 40 =(2n + l)rt-30, cos 40 = - cos30, => 2cos220 -1 =-(4cos30-3cos0), => 2(2xz -1) -1 = - (4x3 - 3x), where x = cos0, 8x4 + 4x3 - 8x2 - 3x + 1 = 0, , 4 sin2 0 + 2 sin 0-1=0, =>, , A=B, a2 = b2 + c2, , 87., , '0, 7 1 - tan2, 6 x2 tan, .2, +---------0, 1 + tan2, 1 + tan2, 2,, 2j, , =>, , A2-2A.B + B2 = 0, , =>, , Now 7 cos 0 + 6 sin 0, , 7-7 tan2, , 6, , Hence, the maximum value of tan A tan B, 7C, will, = tan — tan— = -=•-= = 6, 6 V3 V3 3, 86. Let3? = A and 3^2-2=b, , 2 tan, 2J, =1, +---------(1 + tan2, , e, , A=B = ^, , =>, , 0A, , - 2 tan, , tan(A + B) = tan I — I = '3, \3/, tan A + tanB, r, ---------------— =V3, 1 - tan A tan B, , =>, , 0, 2, , ■ (0>l, A, 0, 0, 0, => cos, + sin, -sin — < 0, >0 and cos, 2, 2, 2, k27, , tan, , 113, , (x + l)(8x3 - 4x2 - 4x - 1) = 0, Pmn=m^cOix, , (sin x)+nlogCOSJ (cot x), , > ndog^ J(sin x) + logco5 x (cot x)) V m > n, = n0°gco1x. (sin x • cot x)), = n10gco*x. cos x=n, Thus,Pmn>nV m^n, , 89. Clearly, P4,|, , Similarly PM, , 4log±, 41, , + 9 log ! (1) = 4, , 41, , =, , www.jeebooks.in, Mean proportional of P4,, , j and PM, , j is ^9x4 = 6

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www.jeebooks.in, 114, , Textbook of Trigonometry, , 90. PM(x) = P22(x), , Given, , => 3 logcos, :os x, x (sinx) + 4 logro$ x (cot x), = 2(*ogcosr (sin x) + logros x (cot x)), => Xlog,co, x (sin *) + l°gc<:osx(cot x))+ logCOJX (cot x) = 2, 3 + logco, X (cot x) = 2, =>, logco, X (cot x) = - 1, 1, cos X, COt X = (cos x)“ ’, sin x COS X, • 2, 1 - sin, x = sin x => sin 2 x + sin x - 1 = 0, , Number of solutions in ----- , 4n isfc = 5., 2, 94. k = 5, , cos2x -1, sin2x, , 2 sin2 x = |tanx| = -^, 2 cos x sin x, , 96. Sum of interior angles = (k - 2)n = 3n, 97. Now, (2 sin x - cos x) (1 + cos x) = sin2 x, , sin x =, sin x =, , 1, 3 . ., sinx = - or -(rejected), 2, 2 ', , 95., , cos2 x = sin x, , 1, =4, (1 - sin x)2, , S=4, , 2, , 75-1, , (v sin x £- 1 )), , 2, , p + g=7, Thus,, So/. (Q. Nos. 91 to 93), 70 =(2n + l)n, n = Oto 6, 40 =(2n + l)7t — 30, cos 40 = - cos30, => 2cos220 - 1 =-(4cos30-3cos0), , => (1 + cos x) [2 sin x - cos x - 1 + cos x] = 0, =>, (1 + cos x) (2sin x - 1) = 0, 1, cos x = -1 or sin x = 2, 1, So,, sin a = 2, , n], as 0 < a < - j, , V3, cos a = —, 2, 3 cos x - 10 cos x + 3 = 0, , => 2(2x2 - 1) - 1 = - (4x3 - 3x), where x = cos©, cos x = -, cos x * 3, 3, , 8x4 + 4x3 - 8x2 - 3x + 1 = 0, , (x + l)(8x3 — 4x2 — 4x — 1) = 0, o-f ti, . roots, . are cos—, n , cos—, 2n ...,cos---13it, 97., The, 7, 7, 7, it, 13k, 3n, lln, 5n, 9n, where cos— = cos---- , cos— = cos---- , cos— = cos—, 7, 7, 7, 7, 7, 7, , .'. The roots of8x3 - 4x2 - 4x + 1 = Oare cos—, cos—, cos—., 7, 7"7, 4, 8, 4, oo sec—, n .sec—, 3n .sec5n, ; ofr, 9z., —are roots, + 1=0, ,3, X2, 7, 7, 7, x, x, =>, x3 - 4x2 - 4x + 8 = 0, , 7t, 3n, 5n, sec — + sec — + sec — = 4, 7 7, 7, 2 7t, 93. sec —, sec7, - -^-.sec2 — are roots of/(-7x) = 0 •, 7, , (V^)3 - 4(7x)2 - 47^ + 8 = 0, x3 —24x2 + 80x-64 = 0, 2k, 2 3k, 2 5k, sec — + sec — + sec — =24, 7, 7, 7, So/. (Q. Nos. 94 to 96), Let, S = 1 + 2sinx + 3sin2x + 4sin3x + ..., 2, , =>, , cos p =, , 3, 3, and, 1 - sin 2x = cos x - sin x, => sin2 x + cos2 x - 2 sin x cos x = cos x -sin x, =*(cos x-sin x) (cos x-sinx- 1)= 0, 1, sin x = cos x = -7=, V2, or, cos x - sin x = 1, cos x = 1, sin x = 0, cos y = 1, sin y = 0, , Ji, , i, , 3-73+8, , 2, , 3, , 6, , cos a + cos P + cos y = — + - + 1 =, ’, , 98. sin(a - p) = sin a cos p - cos a sin p, , 1, , 1, , Ji, , 1-276, , 2 Ji, , = - X------------ X-------- =--------------, , 2, , 3, , 2, , 3, , 6, , 99. (A) If M is mid point of PQ, then M =, , 0+, 2, , sin 0 + sin <f>, 2, , 3, , sinx.S = sinx + 2sin x + 3sin x + ..., (6, sin 9), , (l-sinx)S = l sinx + sin2x + ..., (1 -sinx) 5 =, , N, , Q (<t>, sin <$>), , P., , 1, , M, , 1 - sin x, , y= sin x, ---------- -X, , www.jeebooks.in, 0, , 5 =----------- -2, (1 - sinx)2, , L

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www.jeebooks.in, Chap 01 Trigonometric Functions and Identities, , Also,, , 0+0, 0+0, N = ------ , sm, 2, 2, , <3 +, , It is clear from the figure., ML<NL, sin 0 + sin 0, 0 + 01, <sin, 2, 2 J, (0+0, =3, sin 0 + sin 0 < 2 sin, , =>, , 3-7<y<3+7, -4<y <10, X = 10, pi = - 4, =>, X + pi =6, X -pi = 14(R,T), n, . (n, (C) Let y = 1 + sin — + 0 + 2 cos---- 0, 4, 4, , I 2, , = 2 sin, , i, , and, , (sin 0 + sin 0) sin — £ 1 (p, q, r, s, t), 4, (B) v a2 + b2 = (sin 0 - sin 0)2 + (cos 0 + cos 0),2‘, , = 1 + 3 cos---- 0, <4, , = 2 + 2 cos(0 + 0), 2 0+0, = 4 cos, < 4 (s, t), 2, , - 1 < COS I — - 0 I < 1, 14, ), , (C) V 3 sin 0 + 5 cos 0 = 5, =>, 3 sin 0 = 5(1 - cos 0), Squaring both sides, then, 9 sin2 0 = 25(1 - cos 0)2, => 9(1 - cos 0) (1 + cos 0) = 25(1 - cos 0)2, =>, , 9(1 + cos 0) = 25(1 - cos 0) (1 - cos 0*0), 34 cos0 = 16, 8, 15, cos 0 = —, then sin 0 = —, 17, 17, 75 24, 5 sin 0 - 3 cos 0 =-------- = 3 (r), 17 17, Hence, 5 sin 0 - 3 cos 0 = 3, 7 + 6 tan0 - tan2 0, 100. (A)Lety =, (1 + tan2 0), = 7 cos2 0 + 6 sin 0 cos 0 - sin2 0, =7, , 1 + cos 0 ., , 2, , J, , 1 - cos 20, + 3sin 20 2, ' ....., , = 3 sin 20 + 4 cos 20 + 3, - J(32 + 42] + 3 < 3 sin 20 + 4 cos 20 + 3, , <7(32 + 42) +3, , =>, , - 3£3cos( --0 <3, , =>, , 1 - 3 < 1 + 3 cos ( - - 0 I < 1 + 3, , 14, , J, , U, , -2<y<4, X = 4, pi = - 2, =>, X + pi=2,X-pi=6(P, Q), 1, 101. (A) | cot x | = cot x + -—, sin x, , n, Ifo < x < — =>cot x > 0, 2, „, 1, So cot X = cot X +----sin x, , 1 = 0, no solution, . ----sin x, , n, 1, If — < cot X < 7t, - cot X = cot X + ----sin x, 2, , =>, , 2 cos x + —5—= 0, sin x, sin x, , =>, , 2n, 1 + 2 cos x = 0 and sin x + 0 => x = —., 3, , (B) since sin 0 + sin 0 = ^, , • 2, , -Ji, , ), , -(i), , and, cos 0 + cos 0 = 2, ...(ii), (ii) is true only if 0 = 0 = 0 or 2n but 0 = 0 = 0 or 2n do not, satisfy (i), Hence given system of equation has no solution., , -2<y <8 => X =8,pi =-2, X + pi =6, X-pi = 10 (R, S), (B) Let y = 5 cos 0 + 3 cos ^0 + y j + 3, , fi, , K, , -- + 0 + 2 cos ( — - 0, 2 \2, \4, (n, \, f ftn, = 1 + cos---- 0 | + 2 cos|----- 0, I4, 4, , = 1 + cos, , sin 0 + sin 0 < V?, ■, , 115, , ', , = 5 cos 0 + 3 - cos 0----- sin0 + 3, k2, 2, ), , 3>/3, n . 3, = — cos0- — sin0 + 3, 2, ', 2, , . i 7C, , |, , | K, , |, , (C) sin" a + sin — a ■ sm — + a, 13, J, \33, JJ, 2, , = sin2 a + sin2 — -sin22 a = -., 3, 4, (D) tan 0 = 3 tan 0, 2 tan 0, tan 0 - tan 0, tan(0 — <+>) =, 1 + tan 0 tan 0 1 + 3 tan 0, 2, ------------------- Max if tan 0 > 0, cot 0 + 3 tan 0, , www.jeebooks.in, ;. 3 -, , |f 13? f-3^, , \ 13, A 3a/3, < — cos 0-------1 ;in 0 + 3, 2, 2

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www.jeebooks.in, 116, , Textbook of Trigonometry, cos 0 + 3 tan 0, , (Using AM > GM), , 2, , 2, , 2, , (cot 0 + 3 tan 0)2 £ 12 =>, , = - 1 + 4 sin, , j + sin, , + sin (y), , A+ B, , = 2 cos, , k, . n, . C, sin----- sin —, 2, , c, , A+ B, , = 1 + 2 sin, , 4, , n-C, , = 1 + 2 sin, , A-B, , cos, , n+C, , A-B, , 4, , 4, , 2 sin, , k 4 ), 7t + C + A — B, , = 4 cos, , n-A, n, , n-A, , 2, , 4, , n+A, , = 1 + 4 cos, k, , n-B, , sin, , 4, , = 1 + 4 cos, , = 4 cos, , 8, , cos, , cos, , n-B, , 2, , 4, , 4, , sin, , n-C, 4, n+A, , 4, , 103., , 4, , n, , n+ B, , = 4 cos, , n-C, , sin, , 4, , 4, , an, ■ MV, •, (B) sin, — + sin, , n + C+ B—A, , sin, , . (n-B, n-A, sin •-------- sin, 4, 4, , 4, , = 1 + 4 sin, , it -C, 4, , n-C, , = 1 + 4 sin, , cos, , 4, , n-C, , 4, , n-C, , sin, , 4, , n-C, , 1, , it + C + B — A, , sin, , 8, , 4, 8, , it + C, , - cos, , 4, , n-C, , = 2 cos, , 4, , n + C+ A-B, , A-B, , cos, , (•: A+B + C =Jt), , 4, , (v A +B +C = it), , 2 sin, , 4, , it —C, , n+C, , - cos, , 4, , sin, , n-C, , = 1 + 2 sin, , 4, , (n -CA [ (A- B\, (n + C, = 2 cos -------- (cos ------ r- - cos, k 4 J[ I 4 ), I 4, , 2., , -2 cos, , 4, , cos, , 4, , I 4, , n+C, , cos, , = (eos^) + cos(|])-(cos(£) + cos^)), , tan2(0 - 0) <, 102. (A) sin, , n+ B, , sin, , (C)cos+ cos-cos(£], , > 12, , tan(0 - 0), , n + A, , 8, , it - B, , sin, , sin, , 4, , cos, , it, , n-B, , 2, , 4, , It + B, , cos, , cos, , 4, 1, , it - A, , 4, , cos, , n, , n-A, , 2, , 4, , n-C, 4, , 1, , 2B +, 1 + tan — 1 + tan1 + tan —, 2, 2, 2, , ,, . A . B . Cl, = k 1 + sm— sin— sin—, 2, 2, 2., , 2A, 25, 2C, => cos — + cos — + cos —, 2, 2, 2, J,, . A . B . C, = 2 1 + sin—sin—sin —, 2, 2, 2, , 4, , — - sin M, —, , [by using identity], , 104 sina - cosy, sinP cos8, , A+B, , = - 1 + 2 sin, , cos, , 4, , A-B\, , sina - sinp, , =>, , 4, , cosy - cos8, , cos8, , sinP, , n+C, , + 2 cos, , k, , 4, , sin, , n —C, 4, , ), , a-P), 2 J, , 2 cos, , fn-cW fA-Bl, fn + C, + cos, k 4------- I 4 J, k 4, , (using dividend©), , sinp, , = -1 + 2 sin ------ (cos --------, , 2sin, , (•.• A +B +C = n), , = - 1 + 2 sin, , cos8, , n-C, 7C, , 4, 2 cos, , n + C + A -B, , cos, , n +C + B - A, , 8, , n -A, , cos, , 4, , 4, , (tan0 + cot0) -(tan30 + cot30) + tan 45°, , n-C, , sin, , tan9 - tan30 + tan50 - tan70 + tan90, tan0 - tan30 + tan50 - cot30 + cot0, , 4, , = -1 + 4 cos, , 105. Let — = 0 => 100 = —, 20, 2, =>, 20 = 18° or 0=9°, , Now,, , n -A, , = - 1 + 4 sin, , 7t, , [using tan50 = tan45°], , 4, , E = -^—, sin20, , _L_ + !, , www.jeebooks.in, = - 1 + 4 sin, , n, , n -A, , 2, , 4, , sin, , n, , n-B, , 2, , 4, , cos, , n, , n-C, , 2, , 4, , sin60

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www.jeebooks.in, 120, , Textbook of Trigonometry, , 124. We have, sin 0(1 + sin2 0) = 1 - sin2 0, , =>, , 127. tan82-° = cot7-° =, 2, 2, , sin 0(2 - cos2 0) = cos2 0, , cos7-°, 2, , sin7-°, 2, On multiplying numerator and denominator by 2 cos7 | °, we get, , Squaring both sides, we get, sin2 0(2 - cos2 0)2 = cos4 0, (1 - cos2 0) (4 - 4 cos2 6 + cos4 0) = cos4 0, - cos6 0 + 5 cos4 0-8 cos2 0 + 4 = cos4 6, , =>, , 2cos27-°, 1, tan 82- 0 =----------- ?-----2, 2sin7-°cos7-°, 2, 2, , cos6 0-4 cos4 0 + 8 cos2 0 = 4, n, 125. 16 cos 6 - cos —, 8., , ., , -, , 371, 8, , 5TC, , a, , cos 6 - cos —, 8 ., , COS 0 - COS----, , 7jt>i, 8J, , 1 + cos(45° - 30°), sin(45°-30°), , cos 0 - cos---, , cos 6 — cos —, 8 ., , 8., , I, , X, , la, COS 0, , I, , 371, , - COS-----, , 8., , = 16 I cos 0 - cos —, , I, , 8., , I, , 2, , _, , 371, , 8, , 2, , cos2 0 - sin2 —, 8, , = 16 cos 0 - cos —, \, 8, , .2 71, , = 16 [ cos4 0 - cos2 0 + sin2 — cos', I, , 8, , I, , (, , 7, , ., , -—, , 8, , 1, , = 16 cos 0 - cos 6 + -, , I, , 8., , = 16 I - cos2 6 sin2 6 + - | = 16, \, 8j, , 1 -2sin220, , -sin2 26 1, +8, 4, , 16 cos 40, = 2cos40, 8, , 8, , X =2, 1, t, 1, 126.2k cos cos 40° =, sin 20° -Ji cos 20°, -Ji cos 20° + sin 20°, , y/3 sin 20° cos 20°, V3, 1, — cos 20° + - sin 20°, = _2_________ 2, -J3, — sin 40°, 4, sin 60° cos 20° + cos 60° sin 20°, , = 72 + 73 + 5/4 + 76 = (7J3 + •j2)(y/2 + 1), , 2 - m(sin2a + sin2p), , 128. LHS =, , 1 - m(sin2a + sin2P) + m2sin2asin2P, , ______ 2 - 2msin(q + p)cos(a - p)__________, 1 -2msin(a +P).cos(a -P) + 4m2sina cosasinPcosP, , ______________ 2{1 - cos2(a - p)}______________, 1 -2msin(a +p).cos(a ~P) + 4m2sina cosasinPcosp, [using m sin(a + p) = cos(a - p)], _____________________ 2sin2(a ~P)_________ _______ ___, l-2cosz(a ~P) + m22[sin(a + P) + sin(a - p)][sin(a + P) -sin(a-P)’, _______________ 2sin2(a - p)_____, 1 - 2cos2(a - p) + m2sin2(a + P) - m:2sin2(a, , — P), , 2sin2(a - P), 1 - cos2(a - p) - m2sin2(a - p), , 2sin2(a -p), sin2(a - p) - m2sin2(a - p), , 2, 1 - m2, , l/n ., 1, 1, 129. Given tan-(P + y-2a).tan-(y + a ~P)tan-(a + p - y) = l,, 4’, ', ', 4”, 4, where a + P + y = jc, fn-2al (it-2p\ (it-2y>| ,, tan I—-—I tan I—tanl—H = 1, , =>, , (, , aY, , I, , 2A, , PY 1 - tany^, 2, , 2/, , = 11 + tan— 1 + tan-PY 1 + tan-, , sin 40°, , \, , 2 cos 40°, , 5/3 + 1, X a/3 + 1, , 2^2(73 + 1) + (5/3 + I)2 2^2(73 + 1) + (4 + 27?), 2, ~, 2, = -^2(7/3 + 1) + (2 + 5/3) = 76 + y/2 + 74 + 73, , cos 0 + cos —, 8 ., , 2 3n, cos2 0 - cos'>, , 2, , = 16 cos 0 - cos —, V, 8, I, , cos 6 - cos —, 8 ., 8., , x la, cos 6 - cos371, —, 12, , 2^2 + 5/3 + 1, , cos 0 + cos —, , 8., , 2^2, /3-1, 2j2, , 771, , a, , = 16 I cos 0 - cos —, , = 16, , 1 + cosl5°, sin 15°, , =>, , 2,, , 2A, , 2), , a, P, y, a P y, tan— + tan- + tan- + tan—tan-tan— = I0, 2 2, 2, 2, 2, 2, a, P, Y, a, B y, tan— + tan- + tan- = -tan—tan-tan2, 2, 2, 2, 2, 2, , www.jeebooks.in, =>, , 2k2 = 16, , so 18k4 + 162k2 + 369 = 1745, , (i)

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www.jeebooks.in, 127, , Chap 01 Trigonometric Functions and Identities, , <59., , 4, 1, 3, ,22, = COS X - COS X + - + -, , =7 (sin* x + cos* x), where x e R and k > 1, , A:, Now, f< (x) - f6 (x), = i (sin4 x + cos4 x) - | (sin6 x + cos6 x), , 4, , ,2, , 3, +2. ', 4, , 2, , 1, , ...(0, , COS X-----, , where,, , = - (1 -2sin2x- cos2x) - - (1 - 3sin2 x • cos2 x), 4, 6, 1, 1, =---- —, 4 6 12, , 160. Given expression is, tan A, cot A, —-------- +, 1 - cot A 1 - tan A, , 4, , (, , 1, , 2, , Yu4, , 0 < cos x - -, , (ii), , 2.J, , k, , -<A^1, 4, 4, 163. cos(a + p) = - => a + p e 1st quadrant, 5, , sin A, sin A, -------x, sin, A, - cos A, cos A, cos A, cos A, +------ x, sin A cos A - sin A, , and, , sin(a ~P) =, , Now,, , sin3 A - cos3 A, 1, cos A sin A, sin A - cos A, , 3 5, 56, 4 + 12 .56, 33, , tan(a + P) + tan(a - P), 1 - tan(a + P) tan(a - P), , t-TI"4 12, , 3, , 164. cos (P — y) + cos (y - a) + cos (a - P) = —, 2, => 2 [cos (P - y) + cos (y - a) + cos (a - P)] + 3 = 0, , => 2 [cos (p - y) + cos (y - a) + cos (a - P)], , = 1 + sec A cosec A, , + sin2 a + cos2 a + sin2p + cos2p + sin2y, , 161. Given A APQR such that, , + cos2y = 0, , —(0, , => (sin a + sin p + sin y)2 + (cos a + <cos p + cos y)2 = 0, , -(ii), , It is possible when,, sina + sinP + siny = 0, and, cosa + cosP + cosy = 0, Hence, both statements A and B are true., , On squaring and adding Eqs. (i) and (ii) both sides we get, (3 sin P + 4 cos Q)2 + (4 sin Q + 3 cos P)2 = 36 + 1, , => 9 (sin2 P + cos2 P) + 16(sin2 Q + cos2 Q), + 2x3x4 (sin P cos Q + sin Q cos P) = 37, , 24[sin(P + Q)]=37-25, , 2a =(a + P) + (a-p), tan2a, , sin2 A + sin A cos A + cos.22 AA, sin A cos A, 1 + sin A cos A, sin A cos A, , 3 sin P + 4 cos Q = 6, 4 sin Q + 3 cos P = 1, , => a - p e 1st quadrant, , 165., , Area = - x Base x Altitude, 2, , sin (P + Q) =, , Since, Pand Q are angles of &PQR, hence 0° < P, Q < 180°., =>, P + Q = 30° or 150°, =>, P = 150° or 30°, Hence, two cases arise here., Case I, P = 150°, P = 150° => P + Q = 30°, 0 < P, Q<30°, sin P <, =>, , x, , x cos 8, , =>, , 3 sin P + 4 cos Q < — < 6, 2, => 3 sin P + 4 cos Q => 6 is not possible., , X COS0, , = - x (2x cos 9) x (x sin 0) = - x2 sin 29, 2, 2, [since, maximum value of sin 29 is 1], , cos Q < 1, , 3, 3 sin P + 4 cos Q < - + 4, , -----H—, , Maximum area = - x2, 2, 166. Given,, , 1, cos x + sin x = 2, x, , 1 - tan2 —•, 2 tan 2, ---------- 2- +, 2 X, 2 X, 1 + tan2 - 1 + tan2 2, 2, , Case II, P=30°, Hence, R = 30° is the only possibility., , 1, , 2, , www.jeebooks.in, 162., , A = sin2 x + cos4 x, , A = 1 -cos2x+ cos4x, , Let, , 1-t2, x, tan — = t =>, 2, T7?, , 2t, , +, , , *2, , 1, 2

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www.jeebooks.in, 128, , Textbook of Trigonometry, , 2(1 - t2 + 2t) = 1 + t2 =>3f2 — 4t — 1 = 0, , =>, , =>, , As, , t=, , 2± y/7, 3, , n, „ x n, 0<x<n =$ 0< —< —, 2 2, , So, tan, , Now,, , Qi = 1, tan (P, —+—, \2, 2J, P, Q, tan — + tan —, —2, 2=1, ,1 - tan —, p tan —, Q, 2, 2, b, , is positive., x 2 + V7, t = tan — =, 2, 3, „, x, 2 tan —, 2t, tan x =-------- 2, 2 x 1-t2, 1 - tan1 —, 2, , —— = 1, a, , -*=i-‘, a, a, -b =a - c, c =a + b, , =>, , 2, =>, , 3, , -3 (2 + V7), 1 + 277, , tan x = -, , ZK = —, 2, , n, ZP+ZQ = 1 - 2>/7, I-2J7, , ZP k, 2, 4, 2, P, 7C, Q, tan — = tan, 2, 4 2, , =>, , (4 + J7, \, , 167., , Since,, , 2, , 1 -, , tan x =, , Alternate Solution, , 3, , tan x =, , 3, , P, Q are the roots of equation, Since, tan — and tan —, 2, 2, ax2 + bx + c - 0, , and, Also,, =>, =>, , [from Eq. (i)], , P + .tan —, Q - —b, tan —, 2, 2, a, P, Q c, tan — tan — = 2, 2 a, P Q R n, 2 2, 2 2, P-t-Q^n R, 2, 2 2, P + Q 71, 2, 4, , (i), , P, P, tan — + tan —, 2, 2, P, tan —I2, , tan — - tan —, 4, 2, n, Q, 1 + tan — tan —, 4, 2, Q, Q, tan — = 1 - tan —, 2, 2, Q, P, Q, tan — = 1 - tan — tan —, 2, 2, 2, , [v P + Q + R = rc], =>, —4^, , [v, , a, a, -b = a-c, c=a+ b, , = j (given)], , www.jeebooks.in

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www.jeebooks.in, CHAPTER, , Trigonometric, Equations and, Inequations, Learning Part, Session 1, • Trigonometric Equations, • General Solution, Session 2, • Equation of the Form a cos 0 + b sin 0 = c, , • Principal Solution, , • Some Particular Equations, , Session 3, • Solution of Simultaneous Trigonometric Equations • Problems Based on Extreme Values of sin x and cos x, Session 4, • Trigonometric Inequality, Practice Part, • JEE Type Examples, • Chapter Exercises, , Arihanton Your Mobile!, Exercises with the @ symbol con be practised on your mobile. See inside cover page to activate for free., , www.jeebooks.in

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www.jeebooks.in, Session 1, Trigonometric Equations, Principal Solution, and General Solution, Trigonometric Equations, The equations involving trigonometric functions of, unknown angles are known as Trigonometric equations,, e.g., cos0=0,cos2 0-4cos0 = l,, sin2 0 + sin 0 = 2, cos2 0 - 4 sin 0 = L, , sin 0 =, , -Ji, , => 0 = —, , 4, , g _ 71 371 971 1171, 4’ 4 ’ 4 ’ 4 ’, , or, , Thus, the trigonometric equation may have infinite, number of solutions and can be classified as, , (i) Principal solution, , n, , , 1t, , , 3lt, , , 5lt, , 0 =(2n+1) —, n e I, 2, cos 0 = 0, , Thus,, <=>, , A solution of a trigonometric equation is the value of the, unknown angle that satisfies the equation., e.g., , n, , n, , COS0 = O<=>0 = ±—,± ,± ...., 2, 2, 2, , 0=(2n + l)—, nE I, 2, , Result 3 tan0 = 0 <=> 0 = mt, n E I., We know that tan 0 = 0 for all integral multiple of n., tan 0 = 0 <=> 0 = 0, ± it, ± 271, ±3ti, ..., =>, 0 = mt, nE I, Thus,, tan 0 = 0 <=> 0 = mt, n E I, , Result 4 sin0 =sina «=>0 = n 71 +(- l)n a, where nE / and, it it, , ae, , (ii) General solution, , 2*2, 71 71, , Principal Solution, , We have, sin 0 = sin a, where a E---- , —, 2 2, , The solutions lying in the interval [0,2tt] are called, principal solutions., , Now,, , sin0 -sina = 0, , <=>, , General Solution, Since, trigonometric functions are periodic, a solution, generalised by means of periodicity of the trigonometrical, functions. The solution consisting of all possible solutions, of a trigonometric equation is called its general solution., , We use following results for solving the trigonometric, equations., Result 1 sin 0 = 0 <=> 0 = mt, n E I., We know that sin 0 = 0 for all integral multiples of it., sin 0 = 0 <=> 0 = 0, ± it, ± 2ti, ± 37t,..., 0 = mt, n e I, Thus, sin 0 = 0, 0 = mt, ne I., , <=>, , 2 cos, , 0 +a, , 0+cO, , cos, , 2 >, '0 + 0^, , <, or, , or, or, <=>, , or, , 71, , 2, , 2, =0, , sin, , ^1=0, =0, , 2 J, 0-a, , or sin, , =0, , 2, , - (2m +1) —, m E I, 2, , r0-a, , < 2 J = m it, m e I, (0 +a) =(2m + 1)ti, mE I, (0 - a) = 2m 7t, m E I, 0 = (2m + 1) 7t-a,me I, 0 = (2m it) + a, m E I, 0 = (any odd multiple of it) - a, 0 = (any even multiple of it) + a, 0 = n it +(-1)" a, where n E I, , Result 2 cos 0 = 0 <=> 0 =(2n +1) —, n 6 I., 2, , www.jeebooks.in, Thus,, , Tt, , We know that cos 0 = 0 for all odd multiples of —., , sin 0 = sin a, , 7C 71, , <=> 0 = n it +(- l)na, where nE I and a e---2 2

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www.jeebooks.in, 131, , Chap 02 Trigonometric Equations and Inequations, , Note, , sine = 1 <=> sine = sin— <=> 6 = nn + (-1)" —, 2, 2, , (i) sin2 0 = sin2 a, , 1-cos20, , 6 = (4n+1)—, ne/V., 2, Result 5 cos 0 = cos a <=> 0 = 2n 71 ± a, nE I and a e [0, TtJ, We have,, where a e [0,7i], cos0 =cosa,, Now,, cos 0 - cos a = 0, '0 + a, (Q-a, -2 sin, •sin, =0, 2, , <=> sin, , "0 +a', 0+a, , ,0-a, = 0 or sin, =0, \ 2, 0 -a, = n it or------ = nit,nE I, , 2, 2, <=> 0 + a = 2hti or 0 - a = 2n7t, n E I, <=>, 0 — 2nlt -a or0 = 2n7t +a, nE I, 0 = 2n7t ±a, nE I, Thus, cos 0 = cos a, 0 = 2n7t ± a, n E I, where a e [0,71], , Note, (i) cos0 = 1 <=>cos0 = cosO <=>6 = 2nn±0 =>0 = 2nn, (ii) cos0 = -1 <=>cos0 = cosn <=>e = 2nn±n, <=> 0 = (2n ± 1)n => 0 = (2n + 1)n, f0, (iii) sin0 = sinaand cos0 = cos a « sin----- = 0, , V 2 J, , 2, , = nn => 0 = 2nn + a, , <=>, , <=>, , (ii) cos2, , 2, 2, cos 20 = cos 2a, 20 = 2n7T ± 2a, n 6 I, 0 = n7t±a, nE I, 0= cos2 a, 1 +cos 20, , 1 +cos 2a, , 2, 2, <=>, cos 20 =cos 2a, <=>, 20 = 2rm ± 2a, n e I, 0 = nTi ± a, n e I, (iii) tan2 0 = tan2 a, , 1 - tan2 0, , 1 - tan2 a, , 1 + tan2 0, , 1 + tan a, , [applying componendo and dividendo], <=>, <=>, , cos 20 = cos 2a, 20 = 2rm ±2a,ne I, 0 = rm ± a, n e I, , Summary of Above Results, 1. sin0 =0 «=>0 = rm, it, , 2. cos 0 = 0 <=> 0 =(2n +1) —, 2, 3. tan 0 = 0 <=> 0 = nn, rr 71, , Result 6 tan 0 = tan a <=> 0 = rm + a, n E I where, , ae, , <=>, , 1-cos 2a, , 71 71', , 4. sin 0 = sin a <=> 0 = rm +(- l)n a, where a G —,—, 2 2, , 5. cos 0 = cos a <=> 0 = 2rm ± a, where a E [0,71 ], , 2 ’ 2,', (, , 71 It, , We have, tan 0 = tan a, where a E---k 2 2, sin0, , sin a, , COS0, , cos a, , 71 71, , \, , 2 2/, , 7. sin2 0 =sin2 a, cos2 0 =cos2 a, tan2 0 = tan2 a <=>, , Q=rm ±a, 7t, , <=> sin 0 cos a - cos 0 sin a = 0, <=>, sin(0-a)=O, 0 - a = rm,ne I, <=>, 0 =rm +a, nE I, Thus,, tan 0 = tan a, , 0=nn +awhere ae, , 6. tan 0 = tan a <=> 0 = rm + a, where a e -, , 7t 71, 2’2, , Result 7 sin2 0 = sin2 a, cos2 0 =cos2 a, tan2 0 = tan2 a =>, , 8. sin 0 = 1 <=> 0 =(4n +1) —, 2, 9. cos 0 = 1«0 = 2rm, 10. cos 0 = -1 <=> 0 =(2n + l)7t, , 11. sin 0 =sin a and cos 0 = cos a <=> 0 = 2nn + a, Note, (i) In this chapter 'ri is taken as an integer, if not stated, otherwise., (ii) The general solution should be given unless the solution is, required in a specified interval or range., , www.jeebooks.in, 0 = rm ± a

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www.jeebooks.in, 132, , Textbook of Trigonometry, , I Example 1. If sin a, 1, cos 2a are in GP, then find the, general solution for a., , 3, , Since, cos 0 = - - is not possible as; - 1 < cos 0 < 1, « 1, cos 0 = 2, , Sol. Since, sin a, 1, cos 2a are in GP., 1 = sin a cos 2a, 1 = sin a(l -2 sin2 a), 2 sin3 a - sin a + 1 = 0, , (sin a + 1) (2 sin2 a - 2 sin a + 1) = 0, sin a + 1 = 0, , n, rc, cos 9 = cos —, , =>, , (as 2 sin2 a-2sina+1^0), , 3, , 0 = 2n7t ± —, 3, For the given interval, n = 0 and n = 1., =>, 0=^, , sin a =- 1, sin a = sin, , 3, , 3, , 7t, , I Example 4. Solve cos 0 + cos 30 + cos 50 + cos 70 = 0., , 2, , a = mt + (- l)n, , - j, n e Z, 2>, , a=nn +(-1)', , neZ, , Sol. We have,, cos 0 + cos 30 + cos 50 + cos 70 = 0, (cos 0 + cos 70) +(cos 30 + cos 50) = 0, 2 cos 40 ■ cos 30+2 cos 40 • cos 0=0, cos 40 (cos 30 + cos 0) = 0, cos 40 (2 cos 20 cos 0) = 0, , I Example 2. If - sin 0, cos 0 and tan 0 are in GP, then, 6, find the general solution for 0., Sol. Since, - sin 0, cos 0, tan 0 are in GP., 6, , Either cos 0 = 0, 0 = (2n + 1), 7t, , or, , cos2 0 = - sin 0 • tan 0, 6, , cos 20 = 0 => 0 = (2n + 1) —, 4, it, , cos 40 = 0 => 0 = (2n + 1) —, , or, , 6 cos3 0 + cos2 0-1 = 0, , Note that cos 0 = - satisfies the equation (by trial),, 2, (2 cos 0 — 1) (3 cos2 0+2 cos 0 + 1)= 0, =>, , 0 =(2n + 1)-, (2n + 1) -, (2n + 1) —, 2, 4, 8, , cos 0 = ^ (other values of cos 0 are imaginary), , I Example 5. Find the number of solutions for,, F it', sin 50-cos 30 = sin90-cos70 in 0,—., 2, , cos 0 = cos —, 3, , Sol. Here, 2 sin 50 • cos 30 = 2 sin 90 • cos 70, sin 80 + sin 20 = sin 160 + sin 20, sin 80 = sin 160 or sin 160 = sin 80, 160 = nit + (- 1)" 80, , it, , 0 = 2nn ± —, n G Z, 3, , I Example 3. Solve sin2 0 - cos 0 =, , for 0 and write, , when n is even Eq. (i) becomes;, mt, 80 = mt => 0 = —, 8, , the values of 0 in the interval 0 < 0 < 2it., Sol. The given equation can be written as;, ,, 1, , when n is odd Eq. (i) becomes;, , 1 - cos 0 - cos 0 = —, 4, 7, , 240 = nit => 0 = —, 24, , *3, , cos 0+ cos 0----- = 0, 4, , ...(ii), , ...(iii), , it, , /.For the given interval 0,— Eq. (ii) and (iii) gives the, 2, , 4 cos2 0 + 4 cos 0-3 = 0, , solution as,, (2 cos 0 - 1) (2 cos 0 + 3) = 0, , 3, n 1, cos 0 = -, —, 2, 2, , _, , „ it it, , . it it 5it 7it 3n 11k, , 0 = 0,—,— and —, —,—,—,—,---4 2, 24 8 24 24 8 24, .'. Number of solutions is 9., , www.jeebooks.in

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , I Example 6. Solve, , sinx+icosx ., , 2. Never cancel terms containing unknown terms on the, two sides, which are in product. It may cause loss of, the genuine solution., , ’/-, , i+i, , purely imaginary., s'n*+'cos*, , So/. Here,, , 3. The answer should not contain such values of angles, which make any of the terms undefined or infinite., , (1 - i) (sin x + i cos x), , (I-0(1 + 0, sin x + cos x + i(cos x - sin x), , 2, , tan x= - 1 = tan, , 4. Domain should not change. If it changes, necessary, corrections must be made., , 5. Check that denominator is not zero at any stage while, solving equations., , which will be purely imaginary, if, sin x + cos x = 0, =>, , it, , I Example 9. Find the set of values of x for which, tan 3x - tan 2x= 1, ------------------, , 4, , 71, , x = rm---- , is the general solution., 4, , 1 + tan 3x • tan 2x, Sol. We have,, , I Example 7. Find the general solutions of, , tan 3x - tan 2x, , 1 + tan 3x ■ tan 2x, , 2I+I COS X | + | COSX |2 +| COS X I'|3+.-t0“ _4, , =>, , Sol. 2, + | cos x | +1 cos x |2 + |, , 1, , =>, , It, , 1, 1 - I COS X |, , 1, , 1, , x = rm + — [using tan 0 = tan a <=;> 0 = rm + a], 4, But for this value of x,, , =2, , 1, COS X = ± -, , COS X I = -, , =>, , 2o, , tan 2x = tan 2rm + — = °°, which does not satisfy the, , I, , 2, It, , sinx, T" = 0, x, 3x, cos - cos —, 2, 2, and show their solutions are different., , = 2nn ± —, (2n ± 1) 7t + —, 3, 3, , I Example 10. Solve sin x = 0 and, , . n, x = rm ± —, 3, , I Example 8. If x * ~ and (cos x), , sin2 x - 3 sin x + 2, , j, , then find the general solutions of x., Sol. As x # —, , Sol. We have, sin x= 0 => x = rm,, , sin x, =>, , COS X t6 0, 1, — 1, , 2, , (sin x-2)(sin x-l) = 0, =>, sin x = 1,2, where, sin x = 2 is not possible and sin x = 1 does not satisfy, the equation., No general solution is possible., , Some Important Results, , i.e., , i.e., , But, , 3x, , x, , x, 3x, cos — cos —, , (COS x)™2x-3™* + 2=1, , sin2 x-3sin x + 2 = 0, , (0, , x= 0,7t, 2k, 3tt, ..., , i.e., Where as,, , 2, , So,, , 2, , given equation as it reduces to indeterminate form., Hence, the solution set for x is <|>., , x = 2rm ± —, 2rm ± 71---3, 3, , =>, , = 1 => tan(3x - 2x) = 1, , n, tan x= tan —, 4, , tan x- 1, , COS X |3 + ... to °°, , = 22, , - I cos X |, , 133, , = 0, where cos — * 0 and cos —, 2, 2, , 2, x it 3n 5rt, — * —, —, —,... and, 2 2 2 2, _, , 7t, «*, , 3x, —, 2, 5n, , 0, , It 371 571, , ,—, 22' 22 2, , x / it, 3it, 5tt, ... and —, tc, —, ..., 3, 3’, 3, , ,..(u), , sin x, ---------------- = 0 => sin x = 0, x, 3x, cos — cos —, , 2, , 2, , x = fl, 2n, 3ti, 47t,..., =>, From Eqs. (ii) and (iii),, x = 2n, 4n, 6n,..., , ...(iii), , ,(iv), From Eqs. (i) and (iv);, The two equations are not equivalent. Since, some solutions, of the first do not satisfy the second equation., , www.jeebooks.in, 1. While solving a trigonometric equation, squaring the, equation at any step should be avoided.as far as, possible. If squaring is necessary, check the solution, for extraneous values.

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www.jeebooks.in, 134, , Textbook of Trigonometry, , I Example 11. Find number of solutions of, tan x + sec x = 2 cos x in [0, 2ti]., , Case II If cos x < 0, i.e. x e, , Sol. Here, tan x +sec x = 2 cos x => sin x +1 = 2 cos2 x, , - cos x = sin x, , =>, But, , 2, , 1, , it, , 2 sin x + sin x - 1 = 0 => sin x = -, - 1, 2, , tan x =- 1 => x = mt---4, , 3tt, , sin x = --1, , 571 7TC ] ., 7t 3n, u —, — , then, 2* 2, 2 2J, , x = — which does not satisfy, 2, , _, , „ ,, , T_, , 3n 771 1171 1571, 4’4’ 4, 4, , 371 117t, , .’. If cos x< 0, the possible values of x are —,, , tan x + sec x = 2 cos x., , Thus, the possible number of solutions are 4., , 1 7t 57t, , sin x = - => x = —, —, 2, 6 6, Number of solutions of tan x+ sec x = 2 cos x is 2., , Thus,, , I Example 15. Solve cot 0 = sin 20 by substituting, 2 tan 6, sin 20 =, and again by substituting, 1+ tanF, 20, sin 20 = 2 sin 0 • cos 0 and check whether the two, answer are same or not., , I Example 12. Solve sec x -1 = (V2 -1) tan x., Sol. We have, sec x - 1 =(Ji - 1) tan x, 1 - cos x = (Ji - 1) sin x, , =>, , Sol. Method I, 2 sin2 — - (Ji - 1) • 2 sin — cos — = 0, 2, 2, 2, , Put, , sin 20 =, , x, 2sin — ] sin — -(^2 - 1) cos —, =0, , 2 I, , 2, , sin— sin —-(V2-1) cos — = 0, 2, 2, 2, , 1, tan 0, , it, sin — = 0 or tan — = (Ji — 1) = tan —, 2, 2, 8, , 1 + tan2 0, , 2 tan 0, , 1 + tan2 0, , tan2 0 = 1 => tan2 0 =( I)2 = tan 2 n, 4, , — = mt or — = mt + —, 2, 2, 8, „, , 2 tan 0, , 2 tan2 0=1 +tan2 0, , 7t, , X, , 1 + tan2 0, , cot 0 = sin 20 => cot 0 =, , 2, X, , X, , 2 tan 0, , A, , 71, , x = 2nn, 2mt + —, 4, , Method II, , Note, , ,71, , ...(i), , 0 — mt ± —, 4, Put sin 20 = 2 sin 0 cos 0, cot 0 = sin 20, , 71, , 0 * (2n+1)-, otherwise the equation will be meaningless., 2, , c°- — = 2 sin 0 cos 0, sin 0, , cos 0 = 2 sin2 0 cos 0, , I Example 13. Solve tan 0 + tan 20 + tan 0 • tan 20 -1., Sol. We can re-write the given equation as;, tan 0 + tan 20 = 1 - tan 0 ■ tan 20, tan 0 + tan 20, 71, ==>, --------------------- = i, => tan 30 = 1 - tan —, 1 - tan 0 tan 20, 4, , 30 = mt + — => 0= — + —, 4, 3, 12, , I Example 14. Find the number of solutions of, |cos x | = sin x, 0<x<4n., 3tt 57t, 771, n, , Sol. Case I If cos x > 0, i.e. x e 0, —, 2, then, cos x =sin x., 7t, , 2 ’ 2, , O —, 471, , 71 571 97t 137T, , tan x = 1 => x = mt + — =, 4’4’4’ 4, 4, , 2, , cos 0(1 - 2 sin2 0) = 0, 2, , 1, , cos 0 = 0 or sin 0 = 2, , 2, , 1, , = sin, , 2, , 2 7t, , 4, , 7E, , 7T, , 0 = (2n +1)— or 0 = n7i ± —, , (ii), , From Eqs. (i) and (ii), it is clear, first method gives less, number of roots then the second method., Note As far as possible, avoid the use of following, formulae, 2, , 2 tan x, . „, -„, 1 - tan x, sin 2x =--------- -T—, cos 2x =--------- -—, 1 + tan2 x, 1 + tan2 x, , tan 2x =, , 2 tan x, , 2~—, , 1 - tan x, , and tan 3x =, , 3 tan x — tan x, , 1 - 3 tan2 x, , www.jeebooks.in, 71 971, , If cos x > 0, the possible values of x are —,, , As these formulae are not defined for some real values of x., Hence, in many cases the solution obtained with use of, these formulae may not be the complete solutions set of the, given equation.

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www.jeebooks.in, 135, , Chap 02 Trigonometric Equations and Inequations, , Exercise for Session 1, 1. Solve sin 5x = cos 2x., , I, , 2. Find the number of values of x in [0,5tc] satisfying the equation 3cos2 x -10cos x + 7=0., 3. If 2 tan2 x -5secx is equal to 1 for exactly 7 distinct values of x g 0,^ ,neN, then find the greatest value, ofn., , 4. Find the general solution of equation sec',2 x = 72(1-tan2 x)., 5. Solve 7cos2 0+3sin2 0=4., , 6. Find the general solution of the equation tan2 a + 2^3 tan a=i, , 7. Find the number of solutions of sin2 x -sin x-1=0., 8. Find the general values of 0 satisfying tan 0 + tan — + 0=2., , \4, , J, , 9. Find the general solution ofsinx + sin5x=sin2x + sin4x., , 10. Solve cos 0cos 20cos 30=-., 4, , 11. Solve2cot2x -3cot3x =tan2x., 12. Find the roots of the equation cot x-cos x = 1-cotx cos x., 13. If the equation x2 + 4xsin0+ tan 0=01 O<0<— j has repeated roots, then find the value of 0., , k, , 2;, , 14. Find the number of solutions of the equation 2sin3 x + 6sin2 x -sin x -3=0 in (0,2n)., 15. Find the number of roots of the equation 16sec3 0-12 tan2 0-4sec0=9in interval (-it,it)., , Session 2, *~»—~rwnr~TM —Tmm~tijbi, , mi im«—iMMncimw, , iT MiiK-nnj WTi—ffwnn- xs tjl~ +■ ~nr.w <i»~m r~u wriM~rr m ~rrrnr_rv, , o >->■, , Equation of the Form a cos 0 + b sin 0 = c and, Some Particular Equations, , Equation of the Form, a cos 0 + b sin 0 = c, To solve the equation a cos 0 + b cos 0 = c, put a = r cos (J), and b = rsin<p, where, r = ^a2 +b2 and (p = tan-1 b, a, Substituting these values in the equation, we get,, rcos 4>cos 0 + rsin (|)sin0 = c, , cos (0 - (?) = - => cos (0 - (p) =, r, Ja2, , c, , + b2, , If | c | > \a2 +b2, then the equation a cos 0 + b sin 0 = c has, no solution., , www.jeebooks.in, If | c | < ^a2 + b2, then put, , IcI, , / 2 , .2, , •ya + d, , cos(0 - 0) =cos a, , = cos a, so that

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www.jeebooks.in, 136, , Textbook of Trigonometry, , 1 + sin x = -Ji cos x, , =>, , (0 - 0) = 2n7t ± a, 0 = 2nn ± a + 0, where n G I, , or, , -Ji cos x - sin x = 1, , On dividing both sides by -Ja2 + b2 i.e. 4, V 4 = 2, we get, , Working Rule, Step I Check whether | c | < 7a2 + b2 or not. If it is, , -Ji, 1.11, — cos x — sin x = 2, 2, 2, , satisfied, no real solution exists., , ., , 71, , 7t ., , 1, , 6, , 2, , cos — cos x - sin — sin x = -, , Step II If the above condition is satisfied, divide both, , 6, , sides of the equation by ya2 + b2., cos, , I Example 16. Prove that the equation, p cos x - qsin x = r admits solution for x only if, , -7p2 + <?! <r<7p2 + <72-, , it, , or, , p cos x - q sin x = r, , Sol. Here,, , „, , it, , 7t, , 6, , 3, , x + — = cos, , , it, , X + — = 2rt7t ± —, 6, 3, „, 71 „, It, x = 2n7t + —, 2nit----6, 2, 7C, , Now, when x = 2nit + —, there are solution for n = 0,1 and, 6, , On dividing both sides by -J p2 + q22, we get, , ------------= cos, COS x, X -, , 77 + <72, 2, , q, , ., , = sin x =, , -Jp2 +q 2, , Put, __ p = = COS 0,, /p^+q 2, , q, , r, , + ?2, , = sin 0 in Eq. (i), we get, , Jp2 + q 2, r, , cos 0 cos x - sin 0 sin x = ., , Jp2+?, , 71, , when x = 2n7t-----, there are solution for n = 1., 2, Thus, total number of solutions are 3., , I Example 19. Prove that the equation, k cos x - 3 sin x = k +1 is solvable only if k g (- <», 4]., Sol. Here,, k cos x - 3 sin x = k + 1, could be re-written as;, , COS (x + 0) = —■=, , =>, , —r, , / _2 J. z,2, , >lP +q, , As we know, - 1 < cos(x + 0) < 1, .".The above equation posses solution only if,, , -1<, , I / =, <11 or - ^p2+q2 <r<ylp 2 + ?2, Jp2+q 2, , k, , jk2 + 9, , or, , 3, , cos X, x ------- 7= =, COS, , -jk2, , COs(x + 0) =, , ., , sin x =, , k +1, , +9, , k+1, , Jk2 +9, , which posses solution only if, - 1 <, , k+1, , <1, , +9, , I Example 17. Solve sin X+ 75 cos x = 72., Sol. Given, -Ji cos x +sin x =-Ji, dividing both sides by, , yla2 +b2 = 7(^3? +12 = 2, we get, Ji, 1 ., -Ji, — cos x + - sin x = —, 2, , =>, , cos, , i.e., 1_, , 7t, , x---6, , 71, , i.e., , Ji, , 2, , 2, , , 7t, , x------- 2n7t ± —, 6, , 4, , , 71, , It, , X = 2n7t ± — + —, 4, 6, 571 n, , it, , k +1, , ■Jk2 +9, , <1, , (fc+1)2 <fc2+9, k2 + 2k + l<k2 +9, k<4, , or, Thus, the interval in which, k cos x-3sinx = fc + l admits, solution for is (-co, 4]., !, , =cosG), „, , i.e., , ,, , x = 2nit + —, 2nit----- , where n G I, 12, 12, , I Example 18. Find the number of distinct solutions of, sec x + tan x = 75, where 0 < x < 3n., , I Example 20. Let [.] denotes the greatest integer less, than or equal to x and f(x)=sin x+cos x. Then, find, / JI \, , the most general solution of f(x)= /I — I ., Sol. Here,, /f ~ 1 =sinl8° + cos 18° = -Ji [sin (45° + 18°)] =-Ji sin 63°., , www.jeebooks.in, Sol. Here, we have sec x + tan x = -Ji, , As sin 63° > sin 45° = -7= and sin 63° < 1, , V2

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , -[ n, , 1<JU, , 10, , So, the equation is sin x + cos x = 1., 7t, cos X-----, , 1, , 7t, , V2, , 4, , „, , n, 71, 0 = n7t + —, 6, , =1, , —(i), , or, , tan0=-V3 = tan, , =>, , _, 7t, 0 = n7t—, 3, , , n, , = -j= or x— = 2nrr ± —, , 4, , 137, , 4, , n „, x= 2nK + —, 2rm, 2, , (ii), , 71, , 71, , 0=nn+— or0 = rm—, 6, 3, , I Example 21. Find the number of solutions of, cos x = 11 + sin x |, 0 < x < 3n, , I Example 23. Solve the equation, 5sin2 x-7sinxcosx + 16cos2 x = 4., , So/. As we know, 1 +sin x > 0, for all x, , So, cos x = 1 + sin x, for all x => cos x- sin x = 1, , On dividing both sides by yja2 + b2 i.e. by V2, we get, , =>, , 71 ., , i, , I, , =>, =>, , =>, , 71, , I, , ft, , I, , 4, , 7t, , 71, , „, , tan2 x-7 tan x + 12 = 0, , , 71, , =cos — => x 4— = 2rm ± —, 4, 4, 4, , n, 71, where 0 < x < 3n, x = 2rm, 2rm —,, 2, 3n, x = 0, —., 2n are the only solution., , Now, it can be factorised as;, (tan x - 3) (tan x - 4) = 0, tan x = 3,4, i.e., tan x = tan (tan-1 3), or, , Thus, number of solutions are 3., , tan x= tan (tan-1 4), , x = rm + tan-1 3, , Some Particular Equations, Equation of the Form, a0 sin'1 x + Qj sin" “1 x cos x + a2 sin1"-2x, cos22 x + ... + an cos" x =0,, , where a0, .....an are real number and the sum of the, exponents of sin x and cos x in each term is equal to n, are, said to be homogeneous with respect to sin x and cos x., For cos x 10, above equation can be written as,, a0 tan" x + aj tan"-1 x + ... + a„ =0, , I Example 22. Solve 3cos2 0-273, , or, , x = rm + tan-1 4, , Equation of the Form, R (sin kx, cos nx, tan mx, cot Zx) = 0,, , where R is a rational function of the indicated arguments, and k, l,m, n are natural numbers, can be reduced to a, rational equation with respect to the arguments sin x,, cos x, tan x and cot x by means of the formulae for, trigonometric functions of the sum of angles (in particular,, the formulas for double and triple angle) and then reduce, the obtained equation to a rational equation with respect, to the unknown, t = tan —, by means of following, , formulae;, , sin0cos0-3sin20 = O., Sol. The given equation can be written as:, 3tan20+273tanO-3=O, =>, , sin2 x - 7 sin x cos x + 12 cos2 x = 0, , On dividing by cos2 x both sides, we get, , cos — cos x - sin — sin x = cos —, 4, 4, 4, cos x + —, , Given equation can be written as, 4(sin2 x + cos2 x) + sin2 x - 7sin x cos x + 12cos2 x = 4, , 1, 1.1, —i= cos x —sin x = —,=, V2, V2, V2, 7t, , Sol. To solve this equation, we use the fundamental, trigonometric identity, sin2 x + cos2 x = 1, , A -2^ ±J12+36 1 r, tan0 =-------- -------- =V3, V3, , 6, , 1 - tan2 2 tan —, 2, sin x =-------- —, , cos X =------------1 + tan2 1 + tan2 —, 2, 2, , www.jeebooks.in, Either,, , tan0=-U= tan—, y/3, , 6, , 1 - tan2 2 tan —, tan x =--------- —-, cot x =------------- —, 1 - tan2 2 tan —, 2, 2

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www.jeebooks.in, 138, , Textbook of Trigonometry, , Equation of the Form, , I Example 24. Solve the equation, (cos x-sin x) 2tanx +, , cos x?, , +2=0, , R (sin x + cos x, sin x • cos x) = 0,, , where R is a rational function of the arguments in, brackets. To solve such equations, put sin x + cos x -1 ...(i), , Sol. Using above formulae, we get, x, 1 + tan2 4 tan —, 2 - +----------- — + 2 = 0, 2 X, X ,, X, 1 -. tan2 —, 1 + tan2 - 1 + tan2 - 1 - tan —, 2, 2, 2, 2, /, 2A, 4t, V - t2, 2t, +2=0, J + t2 l + t2y, „., , , . 2 x, 1 - tan 2., 2 x, , x, , 2 tan 2, , +^J, , (Taking tan — = t), 2, 3? + 6? + 8? -2t -3, , =>, , (t2 +l)(l-t2), , \, , R (sin x - cos x, sin x cos x) = 0, to.an equation;, , 2, , (i4, , x = — cos 2x, 16, , 2, , i5, , 29, , 16, , 2, , =>, , 2, , •lit2 - t - V2 =0, , 1, , whose only real root is, t = -., 2, , =0, , ,, , Thus, the solution of the given equation reduces to the, solution of two trigonometric equations, , 24? - 10? - 1 = 0, 2, , t2 - f, , , we get, , The numbers tx =VI, t2 = —7= are roots of this quadratic, V2, equation., , 1+t, 2, , t-2^2, , 2, , 4, , Put cos 2x = t,, , I 2, , t2 -1, , sin x • cos x =, , = — cos 2x, 16, , 2, , fl-t, , ,, , I Example 26. Solve the equation, sin x + cos x - 2VI sin x cos x = 0, , Sol. Using half-angle formulae, we can represent the given, equation in the form;, , I, , =0, , R t,, , So/. Let (sin x + cos x) = t and using the equation, , 29, , 1 + cos 2x, , =0, , 2, “7, , Similarly, by the substitution (sin x - cos x) = t, we can, reduce the equation of the form;, , I Example 25. Solve the equation, , (T - cos 2x, , ,(ii), , Taking (i) and (ii) into account, we can reduce given, equation into;, t2, , 71, , x + cos, , t2 -1, sm x cos x =-------2, , =>, , R t,, , x = 2nn ± —, is required solution., , sin, , = 1 + 2 sin x cos x, , /■, , Thus, the solution of the equation reduces to that of two, elementary euqations,, 1, X, x, 1 x, => —, = nn ±, -71, —, tan — = —= and tan — =, 2 ^3, 2, 3, 2, 6, , 10, , (sin x + cos x)2 = sin2 x + cos2 x + 2 sin x cos x, , =0, , Its roots are t, = 1 and t2 = - J_, ‘ V3, ‘, VT, , • 10, , and use the following identity, , sin x + cos x = -Ji, , and, , 1, sm x + cos x = —7=, V2, , 1 ., 1, -7= sm X + -7= COS X = 1, , and, , i.i, i, —= sin x + -=■ cos x = - -, , 1, , cos 2x = - => 1 + cos 4x = 1, 2, , or, , , 71, , cos 4x = 0 => 4x = (2n + 1) —, 2, rm n, x = — + —; ne I, 4, 8, , or, , n , . 7t, sin x • cos — + sm — cos x = 1 and, 4, 4, , sin I x + — | = 1, k, , Note, Some trigonometric equations can sometimes be simplified, by lowering their degrees. If the exponent of the sines and, cosines occurring into an equation are even, the lowering of, the degree can be done by half angle formulas as in above, example., , =>, , and, , . ( . nA, sm I x + — I =, , 4J, , and, , 2, , 4, , 4, , I, , 4J, , x + — = (4n + 1) —, 4, 2, , JI, 2, n, ,, ., n, 1, sm x cos — + sin — cos x = —, ■Ji, , V2, , v2, , 1, 2, , Tt, , X + — = Hit + (— l)n •, 4, , www.jeebooks.in, x = 2rm + —, 4, , and, , n, , x = nn + (-1)', , 6, , 4

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 139, , Exercise for Session 2, 1. Solve the equation sin x+cos x=l, , 3, , 2. Solve73cos0-3sin0 = 4sin20cos30., 3. Solve cot 0+cosec 0 = V3., , 4. Solve 42 sec 0+ tan 0=t, , I, i, , 5. Find the general solution of(V3-1)sin0+(5/3 + 1)cos0=2., 6. Find the number of integral values of k for which equation 7cos x + 5sin x =2k +1 has at least one solution., [Hint: a cos 0+b sin 0=c has solution only when \:\<yja2 + b2]., , 7. Solve 2sin2x-5sinxcosx-8cos2x=-2., 8. Solve the equation (1 - tan 0) (1 + sin 20) = 1 + tan 0., , Session 3, r*~—, , ■ 1 'i, , i ii, , .’rti —, , i i n, , Solution of Simultaneous Trigonometric Equations, and Problems Based on Extreme Values of, sin x and cos x, , Solution of Simultaneous, Trigonometric Equations, Here, we discuss problems related to the solution of two, equations satisfied simultaneously., , _, , 1, , A, , it, , 7 it., , ,, , ,, , ,, , i, , tan 0 = -^= => 0 = — or — [value between 0 and 2n], 7 Tl, , Common value of0 = —, 6, 7 TC, , The required solution is, 0 = 2rm + —, , We may divide the problem into two categories :, , (i) Two equations in one ‘unknown’ satisfied, simultaneously., (ii) Two equations in two ‘unknowns’ satisfied, simultaneously., , I Example 27. Find the most general values of 0 which, 1, 1, satisfies the equations sin 0 = - - and tan 0 = ~^=., Sol. First, find the values of 0 lying between 0 and 2n and, satisfying the two given equations separately. Select the, value of 0 which satisfies both the equations, then gener­, alise it., 1 n 7n lift, sin 0 = — => 0 = — or----, , 2, I Example 28. If tan(A-B) = l and sec (A+B} = -j=,, , then find the smallest positive values of A and B and, their most general values., Sol. For the smallest positive values, find A + B and A - B, between 0 and 2n from the given equations., Since, A and B are positive angles, A + B > A - B. Solve the, two to get A and B., For the most general values, find the general values of, A - B and A + B by solving the given equations separately., Solve two to get >1 and B, , www.jeebooks.in, 2, , 6, , 6, , 71, , tan (A - B) = 1 => A - B = — or —, 4, 4

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www.jeebooks.in, 140, , Textbook of Trigonometry, , Also,, , 1171, 71 or-^sec (A +B) = -4==> A + B = —, 6, 6, V3, , I Example 30. If r > 0, - n < 0 < n and r, 0 satisfy, r sin 0 = 3 and r = 4 (1 + sin 0), then find the possible, solutions of the pair (r,0)., , A + B > A - B,, I171, A+B„ =-----, , Since,, , (ii), , 6, , or, , ,, , 24, , —, , 24, , r = 4 (1 + sin 0) and rsin 0=3, , On eliminating 0 from above equations; r = 4 ( 1 + -, , On solving Eqs. (i) and (ii), we get, 25n _ 197C, A—, , Sol. Here,, , I, , 24, , r2 - 4r - 12 = 0, , 24, , (r -6)(r + 2) = 0, r=6orr = -2, , 1, 3, r sin 0 = 3 => sin 0 = - or sin 0 = —, 2, 2, , Solve for the most general values, tan(A - B) = 1 => A - B = nit + —, , ...(iii), , r, , .-, , ,, , ., , 3, , 7T 571, , 2, , 6, , Neglectmg sin 0 = —, we get 0 = —, —, , 2, , -A, , sec (A + B) = -7= => cos (A + B) = —, V3, 2, , (r, 0) =, , I 6), , A +B = 2m7t ± —, 6, On solving Eqs. (iii) and (iv), we get,, , 66, , are the required pairs., , Problems Based on Extreme, Values of sin x and cos x, , A. = -1 (2m + n)n + — ± —, 4, 6, 2, 1 \, , I, , 6, , ...(iv), , 4, , „, , 5tt, f 6, —1 and ^6, —, , 7t, , 7t, , B = - (2m - n)n-----±— where m, nel, 2 ', 4, 6, , j, , I Example 29. Solve the system of equations, 2n, , sin x, x + y = — and------- = 2, 3, siny, Sol. Let us reduce the second equation of the system to the, form,, sin x = 2 sin y, —(i), ., 2n, 2n, Using x + y = — we get, sin x = 2 sin ------ x, 3, , _( ., , 271, , 271, , I Example 31. Solve2cos2 - -sin2 x = x2 + —,, 2, x, 0<x<—., 2, Sol. In this problem, terms on the two sides of the equation are, different in nature., LHS is in trigonometric form, whereas RHS is in algebraic, form. Hence, we will used inequality method., , Here, LHS = 2 cos2 — ■ sin2 x, 2, = (1 + cos x)sin2 x < 2, , sin x= 2 sin---- cos x - cos----- sin x, I, 3, 3, , [vl + cos x <2, sin2 x<l], , (^3, 1 ., = 2----- cos x + - • sin x, , and RHS = x2 +, , sin x=y/i cos x + sin x, , Hence, LHS is never equal to RHS, .•.The given equation has no solution., , I2, , 2, , cos x=0, , >2, x2, , [v A.M. > GM], , I Example 32. Solve sin6 x = l + cos4 3x., , x = (2n + 1) —, 2, , Sol. LHS = sin6 x < 1, , 7t, , RHS = 1 + cos4 3x > 1, , x = — + rm, 2, , Hence, sin6 x = 1 + cos4 3x is possible only when., , Substituting in x + y = —, we get, 7t, , LHS = RHS = 1, =>, sin6 x = 1, , 6, , sin2 x = 1, , y =— rm + —, , and 1 + cos4 3x = 1, , and cos4 3x = 0, , www.jeebooks.in, x = — + n7t, y = — -rm, where nel., 2, 6, , 2, , =>, , cos x = 0, , and cos 3x = 0, , =>, , cos x = 0, , and cos 3x = 0

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, Now,sin x +cos x = 41, , x = (2m+ l)y, , and, , Tt} <, cos x---- --- 1, , 4J, , 3x = (2n + 1)—, where m, n e I, 2, , =>, , It, , Tt, , 2, , 6, , n, , =>, , x = (2m + 1)— and x = (2n + 1)—,, , and, , x = (2n + 1)—, , cos x — = -1, , I, , 71, , and, , Common values of x is (2n + 1)—, where n el. The required, 2, solution,, , 4J, , x = (2n ± l)7t + —, 4, 1+sin 2x = 2, , 6, , jr, , 71, , 2x = mt + (- l)n —, 2, x = — +(- l)n ...(iii), 2, 4, The value in [- Tt, Tt] satisfying Eqs. (i), (ii) and (iii) is, 7t -3n , ,, „ v, x = —,----- (when n = 0, -1)., 4 4, =>, , I Example 33. Solve sin4 x = 1+tan8 x., Sol. LHS = sin4 x < 1, RHS = 1 + tan8 x > 1, , =>, , (ii), , sin 2x = 1 = sin —, 2, , x = (2n + 1)—, n e I, 2, , =>, , (0, , f 711, , It, , x=(2m + 1)-, , Tt, , x = 2mt + —, 4, sin x + cos x = -41, , and, , where m, n e I, =>, , 141, , LHS = RHS only when, sin4 x = 1 and 1 + tan8 x = 1, , I Example 36. Find the most general solutions for, 2^05 x _21-l/^2, , 2Sinx, , sin2 x = 1 and tan8 x = 0, , Sol. As we know, AM > GM, , which in never possible, since sin x and tan x vanish, simultaneously., Therefore, the given equation has no solution., , osin x ., , x, , /------■ 2C0SX, , 2, , (i), , v, , Now, Eq. (i) admits minimum value when, sin x + cos x is (— 41), , I Example 34. Solve sin2 x+cos2 y = 2 sec2 z., Sol. LHS = sin2 x + cos2 y < 2, , {using - -]a2 + b2 < a cos x + b sin x < -Ja2 + b2}, , RHS = 2sec2 z > 2, , 2sinx +2, ’^, +2COSX, COSX >2-72, >2-, , Hence LHS = RHS only when,, sin2 x = 1, cos2 y = 1 and sec2 z = 1, 9, , 9, , 2, , cos x = 0, sin y = 0 and cos z= 1, cos x = 0, sin y = 0, sin z = 0, Tt, => x = (2m + l)—,y = nit,z = tit,, , or, , 2sinx + 2c°sx, , or, , 2sinx +2co, :osSx >2~^, , >2-2, , 2, , Thus, the equation holds only when,, sin x + cos x = -^2, , where m, n, t are integer., , cos X— I = -1, k, 4;, , I Example 35. Solve the equation, (sinx+cosx)1+sin2x =2, when - rt<x<7t., , Tt „, x---- = 2n7t±7t, 4, , Sol. We know, - -Ja2 + b2 < a sin 0 + b cos 0 < ^a2 + b2 and, , - 1 < sin 0 < 1., .'.(sin x + cos x) admits the maximum value as 41 and, , x = (2n ± l)7t + —, 4, , minimum value as -72,, , I Example 37. Solve | V3 cos x - sin x| > 2 for, , (1 +sin 2x) admits the maximum value as 2. Also,, , www.jeebooks.in, (±72)2 =2., , .'.The equation could hold only when,, sin x + cos x = ±41 and 1 + sin 2x = 2, , •, , xg[0,4tc]., So/. We know, | 73 cos x - sin x | < ^3 + 1 = 2, , and, , | 41 cos x - sin x | > 2 (given), , ...(i), , ...(ii)

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www.jeebooks.in, 142, , Textbook of Trigonometry, , Thus, from Eqs. (i) and (ii), we must have, | V3 cos x- sin x | = 2, 1 ., — cos x — sin x = 1, 2, 2, , n, . 7t ., cos — cos x - sin — • sin x = 1, 6, , 6, , cos, , cos, , Hence, x G, , 7t, X+—, , =1, , 6, , n, 71 I = .1 or cos x + —, x+—, = -l, , 6J, , 6, , x + — = 0,2n, 471,... or 7t, 3ti, 57t,, 6, _ 117t 237t 5ti 17tc, —,----- , for x e [0, 4ji], 6, 6*66, 117t 2371 571 177tl, 6, , 6’6’, , 6 J, , (e), , =>, -1 = 1 (absurd), At x = 271;, [1 + sin x] = 1 and [1 - cos x] = 0, sin x = 1 + 0, or sin 0 = 1 (as x = 0), 0 = 1 (absurd), A, L 7t, 37t, 1, At X = <0, —,71, --- , 271), [2, 2, J, , we do not have any solution., Now, to check for the values lying between them., ( 71A, Case II (a) When x G 0, —, k 2J, , [1 + sin x] = 1, [1 - cos x] = 0, sin x = 1 + 0, ( Tt'l, But since x G 0, — , sin x 1, ( 2J, i, , 71, , sin x = 1 is absurd when x G 0, —, , I 2, , 1 Example 38. Show that the equation,, sin x = [1 + sin x] + [1 - cos x] has no solution for x e R., (where [.] represents greatest integers function)., Sol, , As we know that the period of sin x is 271, we need to, check the solution for x G [0,27t]., Let us first check at those points on which sin x and cos x, are integer and then for the values lying between them., Case I (a) At x = 0, [1 + sin x] = 1 and [1 - cos x] = 0, sinx = 1+0 = 1 orsinO = 1 (as x = 0), =>0 = 1 (absurd), (b), , (c), , (d), , At x = — ;, 2, [1 + sin x] = 2 and [1 - cos x] = 1, sin x = 2 + 1 = 3 (absurd), At x = 71, [1 + sin x] = 1 and [1 - cos x] = 2, sin x = 1 + 2 = 3 (absurd), , * x=—, 371 ,, At, 2, [1 +sin x] = 0 and [1 — cos x] = 1, „ ,, . 3n, (, 37tA, 3tt, sin x = 0 + 1 or sin — =1 as x = —, 2, 2, , I 71, , I, , (b) When x G I —, 7t I, , [1 + sin x] = 1, [1 - cos x] = 1, sin x = 1 + 1 = 2 (absurd), f 3ti, (c) When x G I 7t, — I, [1 + sin x] = 0, [1 - cos x] = 1 =>, sin x = 0 + 1 = 1, i, 3ti i, But x G 7t, — in which sin x, , I, , 2), , 1, , , . ,, , ,, f 3n, sm x = 1 is absurd, when x e it, •—, , I, , 2, , i 3ti j, (d) When x G —, 2ti, V 2, J, [1 + sin x] = 0, [1 - cos x] = 0, sin x = 0 + 0 = 0, r 3ti a, But sin x * 0 when x G —, 2n, , I2, , J, , Thus, the given equation does not posses any solution for, x G [0, 2ti] or in general, sin x = [1 + sin x] + [1 - cos x], does not posses any solution for x G R., , www.jeebooks.in

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 143, , Exercise for Session 3, , 1, 1. Find the general values of 0 which satisfies the equations tan 0=-1 and cos 0=-y=., , '2, , 2. Find the most general solution of cosec x = -2 and cot x = 73., 3. Find the common roots of the equations 2sin2 x + sin22x =2 and sin 2x +cos2x = tanx., , - sin-2 A + sin-2 8 = - (73 -1)., 4. Solve the equations, 73 sin 2A =sin 2B and 73, 2'”, '■, 5. Find the number of solutions of sin2 x cos2 x = 1 + cos2 xsin4 x in the interval [0,7t]., , 6. Solve: 1+sin x sin2 — = 0., 2, 7. Solve: cos 50 x-sin50x=t, 8. Find the number of real solutions of the equation (cosx)5 + (sinx)3 =1 in the interval [0,2ti]., 9. Find the number of solutions of the equation, ,A_ ., 1+eax2x =^2|sinx|-1 + -1-cos2x,, —forxe(0,5n)., 1+sin x, , 10. Find the number of solutions as ordered pair (x,y) of the equation 2s;ec, ', , 2x, , + 2c««2y =2cos2x(1-cos22y)in, , [0,27il, , Session 4, Trigonometric Inequality, Trigonometric Inequality, An inequality involving trigonometric function of an, unknown angle is called a trigonometric inequality., , ., , Solution of Trigonometric Inequality, To solve the trigonometric inequation of type /(x) < a, or, f(x) > a where /(x) is some trigonometric ratio, the, following steps should be taken:, 1. Draw the graph of /(x) is an interval length equal to, the fundamental period of /(x)., 2. Draw the line y = a., 3. Take the portion of the graph for which the, inequalities satisfied., 4. To generalize, add nT(ne T) and take union over the, set of integers, where T, is the fundamental period of, , I Example 39. Find the solution set of inequality, 1, sin x > —., 2, , Sol. When sin x= -, the two values of x between 0 and 2n are, 2, 71, 5ti, — and —. From the graph of y = sin x , it is obvious that, 6, , 6, , 1-71, , 57t, , 6, , 6, Y, , X, , Z7\, , 1/2-/, , —3k, , '-2k -i, , sin x> - for — < x < —, 2, , 1-., i, , . \ 3n/2, i\, , Jt, , •, , 6 2 6 \1/, , !k, , 3;, , r4rt—, , www.jeebooks.in, /(*)•, , Hence,, , 1, sin x > 2

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , or, or, , 3n, -n< x <-----4, 3n, — < x < it, 4, 3it, x e - n, - —, , I Example 46. Solve tan3 x+3>3tanx+tan2 x, , Sol. tan3 x-tan2 x + 3-3tanx>0, tan2 x(tan x -1) - 3(tan x -1) > 0, , ]“[4;, , 3n, u —,n, 4, , I Example 45. Solve sin2x>V2sin2 x+(2-V2) COS 2 X, Sol. sin2x>5/2sin2 x+(2--^) cos2 X, =>, , 145, , 2sinxcosx>72sin2 x+(2-72)cos22 x, tan2 x-fz tanx+(72-l)<0, , (tanx-l)(tanx-(72-l))<0, , (72-l)<tanx<l, , (tanx-l)(tan2x-3)>0, , (tan x - l)(tan x + 73)(tan x - 73) > 0, =>, , (y-l)(y+73)(y-73)>0, where tan x=y, , Sign-scheme of above inequality is as follows:, (-), (*), (-), (*), -------- 1------------ 1------------ 1--------V3, 1, V3, , .'., , —f3<y<lor y>>/3, , =>, , -73<tanx<l or tanx>73, , For -n/2<x<n/2, -n/3<x<x/4 or n/3<x<n/2, /.General solution is, (, n, it, rt, it, x e nn+—,rm+~, rm—,rm+~, I3, 7, 3, 4, , n/8<x<n/4, (, it, n} ,, „, xe nit+—, nit+— where neZ., I, 8, 4J, , where nGZ., , Exercise for Session 4, 1. If 2 cos x <73 and x 6[-n,nJ then find the solution set for x., , 2. Find the set of all x in the interval [0, n] for which 2 sin2 x -3 sin x +1>0., 3. Ifcosx-sinx>1and0<x <2n, then find the solution set for x., 4. Solve sin 0+73 cos 0> 1,-n<0<n., , 5. Find the set of values of x, which satisfy sin x cos3 x >cos x -sin3 x, 0 < x <2it., 6. Find the set of all x in (-y-,, 7. Solve sin4f —, , 13., , which satisfy 14 sin x -11 < 75., , Jx^ 1, , I3J 2, , 8. Solve tan x - tan2 x >0 and 12sin x| < 1., , www.jeebooks.in

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www.jeebooks.in, J EE Type Solved Examples:, Single Option Correct Type Questions, , (a) 0, (c) 4, , sinx +Vicos2 x = 4k, , =>, , • Ex. 1. The number of solutions of equation, 8[x2 -x] + 4[x]=13+12[sinx],[.] denotes GIF is, , Now as |sinx + Vicos2 x| < |sin x|+Vi |cos22 x, , (b) 2, (d) 6, , k = 0 is only possible value., sin x 4-Vi cos2 x = 0, , Sol. (a) 8[x2 - x] + 4[x] = 13 + 12[sinx], , Visin2 x - sin x - Vi = 0, , LHS is always even and RHS is always odd. Hence, no, solution., , -1j=, y/2a;, smx = —, V2, , =>, , • Ex. 2. The number of ordered pairs (x, y) satisfying, , sin x = -, , ^nx2', , (b)2, (d)4, |x|,|y|e[0,2], , sin, , Slxf, , 57t, , • Ex. 4. The general solution of, sin20sec0+V3 tan0=0 is, , =1, , < 3 >, tlx2, .,, ,7t, ----- = (4n + 1)—, 3, 2, , 71 n, , (a)0=mt+(-1)"+' —, 0 =nn; n&l, 3, (b) 0=nn,nG/, , x2=(4n + l)2, , (c) 0 =—,ne/, 2, , 3, v |x| e [0,2], then only possible value of x2 is 2, , (d)0=n7t+(-1)n+’^,ne/, , 4'm=2- JI, , Sol. (b) sin20sec0 + V3tan0=O, , Hence, total number of ordered pairs is 4., , sin0(sin0 + V3)sec0=O, sin 0=0, , • Ex. 3. Number of solutions of, Tt, , „, , x = 2nn---4, Hence, for x G [- 271,27t] we have 2 solutions., , So/. (d)|x| + |y| = 2, , Also, , 7t, , x = 2hti---- or 2nn + —, 4, 4, From Eq. (ii) we can say that the only solution possible is, , k, , (a) 1, (c) 3, , (•.• sin x * JI), , /i, , „, , = 1 is/are, , |x| + |y| = 2 and sin, , 1_, , 0=nn, nel, , f~~, , cos2 — (sinx +V2cos2 x), 4, , (•.• sin0 - y/i,secQ *0), , • Ex. 5. The number of solutions of the equation, - tan2 x + —tan2x =1, xG [- 2it, 2jc] is, 4, , sin|, , (b) 2, (d) 8, , (a) 1, (c)4, , (a)0, (c) more than 2, , Sol. (b) The solution is only possible, when, cos2 —(sin x + Vi cos2 x) j = 1, , 4, , and, , J, , 7t, , 2, , 1, , „, , tan2 x + — tan x =0, 4, J, , | = x2 -2Vix + 4 is, (b)2, , (d)1, tty, , ...(i), ,..(ii), , Sol. (d) We know that -l<sin—^=<1,, 2V3, therefore, we must have, -1<x2-2>/3x+4<1, , =>, , -l<(x-Vi)2 + l<l, -2<(x-Vi)2<0, , www.jeebooks.in, Let’s solved Eq. (i), , —(sin x + Vi cos2 x) = kn, 4, , =>

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www.jeebooks.in, 147, , Chap 02 Trigonometric Equations and Inequations, , But, square of a real number cannot be negative, therefore,, we must have (x-73)2=0, , x=fi, , =>, , Note that x=V3 satisfies the given equation., , • Ex. 8. Number of solutions of the equation, cos4 2x + 2sin2 2x=17(cosx + sinx)8,0<x<27t is, , (b) 8, (d) 16, , (a) 4, (c) 10, , Sol. (a) Let sin2x = y, then 1 + y4 = 17(1 + y)4, , • Ex. 6. x, and x2 are two solutions of the equation, , y = sin2x = - ^ is the only possibility, , Clearly,, , ex cos x =1. The minimum number of the solutions of the, x = 105°, 165°, 285°, 345°., , equation ex sin x =1, lying between x} and x2 can be, (b) 1, (d) None of these, , (a) 0, (c) 3, , (, , Sol. (b) We have ex cosx = 1 or cosx = e~x, , satisfying the equation (5/3)sec2 9 = tan4 0 + 2 tan2 0 is, , Consider the function, f(x) = cosx - e~x, , Clearly, /(x) is continuous in [xp x2] and differentiable in, (xp x2)., , Hence, by Rolle’s theorem, there is atleast one x G (xpx2), , such that f'(x) = - sin x + e~x =0, , sin xex = 1 has atleast one solution G (xp x2)., , • Ex. 7. The product of common differences of all possible, AP which are made from values of ‘x’ satisfying, ‘, , 11, , ’, , 1, , cos2 -Ax +cos, , 2, , (a), , (c), , Sol. (a), , 2, , J, , 471, , 1 + cus(Xx), , 1 + cos(px), , 2, , 2, , 2 cos, , 2, , =>, , (Vs)' = t2 - 1, , => t = 2 is only solution as t > 1, .'.There will be 2 values of0 in given interval., • Ex. 10. Number of solutions of the equation, 7t I, 71, , 7T, , V 2/, , =1, , cos(Xx) + cos(px) = 0, , (X + p)x, , sec2© = t, , 06 0,—, , X-g, , (d) None of these, , =>, , Put, , 4n, , (b), , X2-p2, , = (sec2©)2 - 1 = sec4 0-1, , cot(0)+cot 0+— +cot 0---- +cot(30) =0, where, 3J, \, \, 3y, , 2, , 27t2, , Sol. (a) tan4 0 + 2tan20 = (tan20 + I)2 - 1, , I, , -px =1, 2 >, , X^p2, , (b) 4, (d)1, , (a) 2, (c)0, , /(x,) = 0 = /(x2)., , =>, , • cos, , (W)xk 0, 2, , (b)0, (d) None of these, L Tt'), 7t, Sol. (c) cot(0) + cot 0 + — + cot 0---- + cot(30) = 0, 3, 3, , (a) Infinite, (c) 1, , „ n n, Put 0 = — a, 2, , J, , 71 I, , I, , f, , 2, , 2, , (2n +1)2, , _ (2n + l)7t, , X+p, , (2n + l)7t, or x =----------X-p, , tana + tan —, tana - tan —, tana +---------------- — + ---------------- — + tan 3a = 0, 71, 71, 1 + tana tan — 1 - tana tan 3, 3, 3, , Thus, common difference can be, 2tt, 2tt, ------- or-------X+p, X—p, , o, , 7t, , 7t, , (X-p)x _, , 7t71 |, , tana + tan a---- + tan a + — + tan3a = 0, 3, I 3, , (X + p)x _ (2n + l)^, or, , 71 71, , • Ex. 9. The number of values ofQ in the interval] —, —, <22, , =>, , 3 tana - tan3a, , 1 - 3tan2a, , + tan 3a =0, 7, , 4 tan 3a = 0, tan 3a = 0, 3a = rm, n, a=—, 3, , www.jeebooks.in, Now, product =, , 47t2

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , • Ex. 15. The system of equations, tan x = a cot x, tan 2x = bcos y, (a) Cannot have a solution if a = 0, (b) Cannot have a solution if a = 1, (c) Cannot have a solution if 2>fa > (1 - a)| •, , (d) has a solution for all a and b, Sol. (b,c) If a = 0, then tan x = 0 => x = rm and then for any, value of y such that cosy = 0 the second equation satisfies, option (a) is false., If a = 1 then tanx = cotx => tan2x = l, , • Ex. 17. Which of the following set of values of ‘x’, satisfies the equation, ^Zsin^x -3sinx + I, , tan2x is not defined., option (b) is true, Now from the first equation tan x = 4a, , (c) x = rm, n e I, , (d) x = 2rm + —, n e I, 2, , 1, , \, , [ 2nn72 x ~3nn x+1|, , or, , 12, , (b) -y>“, , 3, , (d) -°°,, D, , O, , 8, u, 3, , 12, 5 ’ °°, , Sol. (a,b,d) cos22 x - 2cos x + 1 = 3 - y + 3, 2y+ 5, _ 5y + 12, (cosx -1)2, ’ 2y + 5, , Also,, =$>, , -1 < cos x < 1 => - 2 < cos x - 1 < 0, 0<(cosx-l)2 <4, , - 4 < 0 =* 2^<0, 2y+ 5, 2y + 5, 3y+ 8, >0 => (3y + 8)(2y + 5)>0, 2y+ 5, 5y +12, >0 =» (5y +12)(2y + 5)>0, 2y + 5, , Now,, =>, , and, , (, , 8', -00,--, , l, , 3J, , u, , 5, -oo,-- U, 2, , and, , 5, 2, , —, , It, , • Ex. 18. ForO <0 < —, the solution(s) of, 2, 6, (, cosec 0 + — 1 = 4V2 is (are), cosec 0 +, 4, k, 4 J, m=1, \, 7, 1 \ n, , (b)5, , (a)7, (0^, 12, , Sol. (c, d) We have,, 6, r, (m - l)n, cosec 0 + — = 44i, \cosec 0 +, 4, 4, , sm —, , 6, , _______ 4_ ------r-------------- i, , S-T- (m - l)n, , m"lsin 0 +, , S-, , =>, , 4, , =>, , sin 0 + —, 4, , sin, , y, , n. > L I, , 0+, , 4, , I, , 510^0 + —^ COS, , 4J, , m «1, , 12, , 4, , = 4, , (m - l)n, 0+— - 0+, 4, t, 4 ) 1, —=4, (m - l)it . f n mn, •" sin|9 +, sin 0 + —, 6, , 00, , —, , 9, , 2sin2 x - 3sin x + 1 = 0, 1, smx = - -,, 2, 1 smx, • =1, smx = -,, 2, , 0<^<4, 2y+ 5, , =>, , 3-^2nn2 x-3an x ♦ 1J, , J, , (1 tin2 x -3nn x*1, , <1, , • Ex. 16. If - --3- = sin2 x + 2cos x +1, then the value ofy, ' 2y +5, lie in the interval, , (a), , ■, , Sol. (a,d) 2[, , 24a <\b(l - a)|, , =>, , 71, , (b) x = rm ± —, n e I, 3, , /, , 8, ,, t + - = 9 => r-9t + 8 = 0 => t = 1,8, t, 2sin2x-3sinx + 1 = 3, , 24a, , 2 tanx, b(l - tan2x), , i =9, , (a) x = rm ± —, n e I, 6, , a must be positive, | cosy | =, , f2- 2sin2 x + 3sin x), , ;+2t, , 2, , Let 2^, , =>, , =>, , 149, , (m - 1)k, 4, , |, , /1 lit I ., , [n, , I, , 4 J, , I, , - cos 0 + — sm 0 +, , (m - l)n, 4, , (m - l)n, sin 0 +, sin 0 + —, 4, I, 4, , 00, , =4, , www.jeebooks.in, ye, , 8, 3, , 5 ’, , 12, —>°°, 5, , =>, , (, (m - l)n, (n, cot 0 +, - cot 0 + —, 4, I, 4, m = l_, \, 6, , =4

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www.jeebooks.in, 150, , Textbook of Trigonometry, , 2rt, n n I, n 2n, cot 0 + — - cot 0 + —, 4, I, 4, , cot0 - cot (0 +, , 5H, +... + [cot^0 + —, , 4, , 6rr, - cot 0 + —, 4, , =4, , ,, B, 1 + 'an 2, , (tan3° + 1) _, , --------------------- — — Id II Zi —, , , . B, 1 - tan —, 2, (, bA, tan(45° + 3°) = tan A = tan 45° 4—, (-tan3° + l), v, ', , V, , cot0 - cot 0 + — =4, , I, , 2J, , A = 48° and B = 6°, , cot0 + tan0 = 4, cos20 + sin:0 = 4 sin0 cos0, , 1---------------, , • Ex. 19. If, , 1 + sin 6°, , cos 6°, , 8 G (0.90° ),.. then, (a) A = SB, (c) A - 78 = 6°, , = tan A =, , '1 + sin 8, , 1 - sin 8, , A can be, (a)110°, (c)300°, , ; where A and, , (b) 260°, (d)190°, -----------•----------A, Sol. (a,b,d), + sin A - fl - sin A = 2 cos —, 2, A, A, .A, . A, A _, AA, sin— + cos----- sin------ cos— = 2 cos —, 2, 2, 2, 2, 2, , (b)8A = B, (d) A + B = 54°, , . B, BA, sin — + cos —, _ . ,, (sin 3° + cos 3° )2, 2, 2, Sol. (a,c,d) -—----------= tan A =, B, . B, (cos 3° -sin*3°), cos-----sin —, Ik, 2, 2J, , A, , 1---------------, , • Ex. 20. If fl + sin A -fl- sin A = 2 cos—, then value of, , 7t, 5n, ( 20 = -1 => 20 = sin, or, 2, 6, 6, 5n, 0 = — or, 12, 12, , =❖, , 27, , 135°, , 45°, , 2, , . B, B, sin3° + cos3°, . Sin 2 + C°S 2, ----------------- = tan A =------4-------- 7cos3°-sin3°------------------- B, . B, cos---- sin —, 2, 2, , So,, , .A, A n, ,, A, . A, sin — + cos — >0 and cos — < sin —, 2, 2, 2, 2, 45° < — < 135° => 90° < A < 270°, 2, , JEE Type Solved Examples:, Passage Based Questions, Passage I, , Sol. (a) (PA) (PB) = (PC) (PD), , (Ex. Nos. 21 to 23), , Consider a circle, in which a point P is lying inside the circle, such that (PA) (PB) = (PC) (PD) (as shown in figure)., , =>, , 4X3= x(2r - x), 12=x(S — x), x2 - 8x + 12 = 0 => x = 6,2, PC, , 6, , PD, , 2, , „, , 22. If PA =|cos0 + sin0 | and PB = |cos0 - sin 0|, then, , On the basis of above information, answer the following, questions:, , 21. Let PA = 4, PB = 3 and CD is diameter of the circle, PC, having the length 8. If PC > PD, then — is equal to, , maximum value of (PC) (PD), is equal to, (a) 1, (b) 2V2, (c)V2, , (d)2, , Sol. (a) PC-PD= (PA) (PB), , = |cos 0 + sin 0| Icos 0 - sin 0|, , www.jeebooks.in, = |cos2 0-sin2 0| = cos 20, , (a) 3, (c)5, , (b)4, (d)6, , Maximum value = 1

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 23. If logpx x = 2, logPfl x = 3,logx PC = 4, then logro x is, , . 0 OA, sin — = —, 2 OP, 0 =60°, , =>, , equal to, =>, , (b)y, , £, , 1, , 10, , 2, , 151, , => — = 30°, 2, , 24. (d) .-. APB is an equilateral triangle., Area =^(5^ = ^, , 4, , 4, A, , Sol. (d)(PA)(PB) = (PC)(PD), , logx((PA)(PB)) = logx((PC)(PD)), logx PA + logx PB = logx PC + logx PD, , ^60°, , . x PD, -11, + - = 4 + Iog, 2 3, 19, logxPD = - - 4 = 6, 6, , B, , = sin 120° + cos 240° + sin 300° + tan 420° + cot 480°, , (Ex. Nos. 24 to 26), , PA and PB are two tangents drawn from point P to circle of, radius 5. A line is drawn from point P which cuts circle at C and, D such that PC = 5 and PD = 15 and Z.APB = 0., On the basis of above information, answer the following, questions:, , 73, , 1, , 2, , 2, , 73, r 1 4-73, — + V3 --;= =---2, , 27, , Let logJ;2 (x + 2) = t, , t=2 + t, t2 -2t -3 = 0, , (b) 25^3, , =>, , 7573, ... 75^3, (d)(c) -----(d) ——, 2---------------------------- 4, , or, , t = 3 or -1, , log]Z2 (x+ 2) = 3, Iog[/2 (x + 2) =-1, 15, x =----- or x = 0, 8, , 25. Value of sin 20 +cos40 + sin50 + tan 70 +cot80 is, => two solutions., , equal to, , 471-1, , (c), , 2V3, , V3, , 26. (c) Given log,z2 (x + 2) = 2 + 31og(x+2,, , 24. Area of AAPB is, , (a), , P, , 25. (b) sin 20 + cos 40 + sin 50 + tan 70 + cot80, , Passage II, , , .2571, w, —, , 60^\, , 3, (b) ___, 4, , Passage III, , 2V3, , 2, (d), , 273, , (Ex. Nos. 27 and 28), , If3sinzx-7sinx + 2 = 0, X6 0,— and/n(0) = sin"0 + cos"0., 2, , 2, , 26. Number of solution(s) of the equation, , logcOJe(* + 2) =2 +3 log (x + 2), , (a)0, (c)2, , ■, , — is, , I2), , (b)1, (d)3, , D, , °\5\[, , <b)il, , (a)H, , / X 65, , (c)^, (c), ii, , no, ., c sin5x + sin4x., 2o. The value of------------------- is, 1 + 2cos3x, , A, '\5V3, e/2>, 'C 0/2>, , 27. The value of fA(x) is, 57, , Sol. (Ex. Nos. 24 to 26), , 5, , On the basis of above information, answer the following, questions :, , P, , (a), (c), , 9, 4^2-2, , (b), (d), , 9, , 472-3, , www.jeebooks.in, OD = OC = 5, PD = DC + CP, CP = 15-10 = 5, OP = 10, , 9, , 9, , Sol. (Ex. Nos. 27 to 28)3sin2x-7sinx + 2 = 0, sin x = ^ or sin x = 2 (reject)

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www.jeebooks.in, Chap 02, , Trigonometric Equations and Inequations, , 153, , JEE Type Solved Examples:, Statement I and II Type Questions, , ■ This section contains 2 questions. Each question contains, Statement I (Assertion) and Statement II (Reason)., Each question has 4 choices (a), (b), (c) and (d) out of, which only one is correct. The choices are, (a) Statement I is true, Statement II is true; Statement II is, a correct explanation for Statement I., (b) Statement I is true, Statement II is true; Statement II is, not a correct explanation for Statement II., , (c) Statement I is true, Statement II is false., , fat , ,, x=—, kG Integer., , =>, 2x=2mt => x=rm, nG Integer, => Statement I is false and Statement II is true., • Ex. 35. Statement I Common value(s) of ‘x’satisfying, , the equations., Iog$inx(secx + 8)>0 and logJinx cosx + logCMX sin x = 2 in, (0,471) does not exist., , (d) Statement I is false, Statement II true., , Statement II On solving above trigonometric equations we, , kjt, • Ex. 34. Statement I x=—,ke I does not represent the, , have to take intersection of trigonometric chains given by, ,, ., it, secx>1 andx=rm+—,nG I, 4, , general solution of trigonometric equation., , Sol. (c) 10gbnxC0SX + 10g£OtJ sinx=2 only, , sin13x - sin13xcos2x = 0, kn, Statement II Both x = m, re I and x =— ,ke I satisfies the, 13, , trigonometric equation., , When sin x=cos x, , it, , x=mt+ —, 4, , => •, , Also, logrinx(secx+8)>0, , sin13x-sin13xcos2x=0, , secx+8<l => secx<7, , =>, , Sol. (d) sinl3x(l-cos2x)=0, , =>, , 7t, , Clearly, x=—+nn satisfy Eq.(ii), 4, Statement I is true and Statement II is false., , sinl3x=0 or cos2x=l, 13x=fcn or cos2x=l, , JEE Type Solved Examples:, Matching Type Questions, • Ex. 36. Match the statement of Column I with the value, , of Column II., , _________, , A, B, , Column II, , Column I, , The number of solutions of the equation, |tan2x|=sinx;x e[0,n], , P, , 1, , 71, , q, , 4, , 71, , The value of 4 tan---- 4 tan3 —, 16, 16, , Sol. (A) -4(q); (B) -> (s); (C) -> (p); (D) -> (p,r), (A) Clearly, number of solutions of |tan2x|=sinx in [0,n], are 4., y= |tan 2x|, , n., 4, , D, , I f the equation tan(p cot x)=cot p( tan x) has, 7t 1, 4, a solution in (0,x)- — k then —, is, 21, n, The value of — in [ 0,2n J if, , n, , y=sinx, , + 6tan2—-tan4 —+1, 16, 16__________________, , C, , 3n, 2, , K, , 2, , r, , 3, , (B) tan 4 A =, s, , 2, , 4 tan A, 1-tan2 A, , 2 tan 2A _, , l-tan22A, 1-, , 2tanA ', , www.jeebooks.in, it, , 5c^2x+2»n2x + 52c«2z+= 126has a, , solution, , =>, , <l-tan2A>, , 4tan A(l-tan2 A), tan4A=--------------------y21 + tan A-6tan A

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www.jeebooks.in, 154, , Textbook of Trigonometry, , = 4 tan A -4 tan’ A +(6 tan2 A - tan2 A -1) tan 4 A =0, , Pmax, , IfA = —, , 71, , 4Pmax, , 4, , n, , _j, , (0) 5COS2 2x4 2 sin 2 x _|_£j2cos2 x+ sin2 2x, , 16, , = 126, , ^.2 2x+ 2 sin 2 x _f_cj2-2sin2 x+l-cos2 2x, , 4 tan — -4tan’ —+6tan2 — -tan1 --1=0, 16, 6, 16, 16, Required value is 2., (C) tan(pcotx)=cot(ptanx), f 71, tan(p cot x)= tan I — p tan x I, , 5”’, , 5<os2 2x4 2sin, 2 sin 2 x, , tjJ-cos2 2x4 2 sin 2 x, , Put cosz2x + 2sin2x = y, we get, 5y + 53'y = 126, , 125, 5y +—=126 => 5V = 125,1 =>y=3, 0, , 7t, , pcotx=nn + —-ptanx, 7t 3n, cosz2x+2sin2 x=3 => x=—,—, 2 2, 2x, cos22x+2sinzx*0 ; — = 1,3, , 71, , nn + —, 71 ., 2, =—smxcosx, P=, tan x + cot x 2, , (vxG[0,7t]), , 71, , Subjective Type Examples, • Ex. 37. lfO<x<3n,Q<y < 3ti and cos x • sin y = 1, then, , =>, , find the possible number of values of the ordered pa'ir(x, y)., cos x ■ sin y = 1, Sol. Clearly,, =>, cos x = 1, sin y = 1, or, cos x = - 1, sin y = - 1, Now, cos x = 1; sin y = 1, „ „, , ,, , o, , j, , » 571, 0,—, 2, , 1, , 55,! 55,-!, , 1, , 2, , 2, , (i), , 371, , -(ii), , • Ex. 39. Find all the values ofQ satisfying the equation,, sin 70 = sin 0 + sin 30 such thatO <0 < 7t., Sol. Given, sin 70 = sin 0 + sin 30, sin 70 - sin 0 - sin 30 = 0, 2 sin 30 • cos 40 - sin 30 = 0, sin 30 (2 cos 40 - 1) = 0, , „ 71, 271, —, , sin 30 = 0 or cos 40 = 2, , 2, , n, 571, 271, ----, , 2, , 371, , „ 371, 3tc, —, 2, , i.e., , i.e. 6 solutions. ', , 71, , 30 = n 7t, n G I or 40 = 2n7t ± —, n G I, 3, Q nTt, nn , 7t, , 3, • Ex. 38. IfQ e [0, 3tc] andreR. Then, find the pairs of, , 2 sin 0 = r4 - 2r2 + 3, , put, , n = 1, 0 = —, 3, , or, , 0=-+^ *, 2 12 2, _ lit 571, u =--- , —, 12 12, „ _ 2lt, n = 2,0 = —, , 2sin0 = (r2 —l)z+2, 2sin0 =(r2 - I)2 + 2 >2, , But max (sin 0) = 1, =>, max (2 sin 0) = 2, /. From Eqs. (i) and (ii);, sin 0 = 1 and r2 - 1 =0, , 2, , 12, , n, 7t, put n = 0,0 = 0 or 0 = ± —, ----- rejected, 12’, 12, , (r,0)satisfying2 sin 0 = r4 -2r2 + 3., Sol. Here,, , 2, , i.e. 4 ordered pairs., , 7t 571, , x = 71, 3ti and y = —, 2, From Eqs. (i) and (ii),, the required of ordered pairs are, , fo,l 2, , and r = ± 1, 2 2, Required ordered pairs are, , \2, , x = 0, 2n and y = —, —, 2 2, and, cos x = - 1, sin y = - 1, =>, , 9 = 5,55, , ...(i), (ii), , n, 12, , www.jeebooks.in, put, , 3

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www.jeebooks.in, 155, , Chap 02 Trigonometric Equations and Inequations, , or, , , 7t I, , „, , .. . 7t, . ,oo V5-1, . sin — = sin 18 =------ , sin, 10, 4, , ., , 71, , 0 = 7t± — n + — rejected, 121, 12, , — ) = sin (- 54°) = -1 - Vs, 4, 10 J, , .’. General solution set, , put n = 3,0 = 7t, „ . ., „ 7t 71 571 771 271 117T, , 7t, = ]0|0 = rm + (- If -1 u ]0|0 = nn +(- l)n —, !, 6, 10, , ,, , .•.Solutions are 0, —, —, —, —, —,---- and 7t., 12 3 12 12 3 12, , I, , Sol. Since, sin 3x > -1 and cos 2x > -1. We have,, sin3x + cos 2x > - 2., Thus, the equality holds true if and only if;, sin 3x = - 1 and cos 2x = -1, 71, , 3x = rm + (- 1)", , i.e., , Ex. 42. Solve the equation, , and 2x = 2rm ± 7t., , 2, , 71, 7t ] ,, ,71, _, x^+(-ir —, and x = rm ± —, n g I, 6, 2, , sin4 x+cos4 x = - sin x-cos x, 2, 7, Sol. Given, sin4 x + cos4 x = - sin x ■ cos x, 2, 7, => (sin2 x + cos2 x)2 - 2sin2 xcos2 x = - sin xcos x, 2, , 7t, 6, , 7, , 1, , :. Solution set is,, x|x = — + (-l)", 3, , 3tc, , u <! 010 = rm -(- 1)" —, i, 10, , • Ex. 40. Solve sin 3x + cos 2x = - 2., , . 71, n x x = rm ± —, 2, , =>, , Note, Here unlike all other problems the solution set consists of, the intersection of two solution sets and not the union of, the solution sets., , 1 — (sin 2x)2 = — (sin 2x), 2, 4, 2 sin2 2x +7 sin 2x - 4 = 0, , =>, , (2sin 2x - 1) (sin 2x + 4) = 0, , or, , • n2x = -1, sin, 2, sin 2x = - 4 < - 1, , (Rejected), , 2x = rm +(- 1)" —, 6, , • Ex. 41. Find all values of 0 which satisfy,, , Sol. Given, sin(30 + a) + sin(30 -a) + sin(a - 9), -sin(a + 0) = cos a, , => 2 sin (30) ■ cos (a) + 2 sin (- 0) cos(a) = cos a, => 2 (sin 30 - sin 0) cos a = cos a, =>, 2 (sin 30 - sin 0) = 1 (as cos a 0 given), =>, 2 • 2 sin 0 • cos 20 = 1, =>, , 4 sin 0 (1 -2 sin2 9) = 1, , =>, , 4 sin 0 - 8 sin3 0 = 1, , =>, , =>, , or, , (2 sin 0 - 1) (4 sin2 0 + 2sin 0 - 1) = 0, 1, - 2± ^4 + 16 -l±7s, sin 0 = - or sin 0 =------- --------- =----------2, 8, 4, Tt, , 0 = rm +(- 1)" —, rm +(- l)n sin, 6, , -1-7T, , rm + (- 1)" sin, , 4, , 9, , Ex. 43. Find all the solutions of, , 4 cos2 x sin x - 2 sin2 x =3 sin x, , Sol. Given, 4 cos2 x sin x - 2 sin2 x = 3 sin x, =>4(1 — sin2 x) sin x - 2 sin2 x - 3 sin x = 0, => 4 sin x - 4 sin3 x - 2 sin2 x - 3sin x = 0, , - 4 sin3 x - 2 sin2 x + sin x = 0, =>, , 8 sin3 0-4 sin 0 + 1= 0, =>, , x = -+(-1)"2, 12, , i.e., , sin(30 + a) + sin(30 - a) + sin(a - 0) - sin(a + 0) = cos a, given cos a £ 0., , - sin x(4 sin2 x + 2sinx-l) = 0, , or, , 4 sin2 x + 2 sin x - 1 =0, , =>, , sin x = 0, , =>, , - 2 ±J4 + 16, sin x = sin 0 or sin x =---------------2(4), x = rm, , or sin x =, , Vs-f, 4, , x = rm, x =rm +(- 1)", , 4, , 71 . f 3n, = sin —, sin 10, I 10, , x = rm +(- l)n 10, I, , 371''l, , 10 J, , General solution set, , /, , = {x|x = zm} ul x|x = rm +(-1)", , www.jeebooks.in, 71, , or, , 71, , Q=rm +(-!)"-, nrt+(-1)" —, 6, 10, -371 1, ., rm + (- l)n ------ , n G I, 10 ), , U- x|x =rm +(- 1)" I - —, k 10

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www.jeebooks.in, 156, , Textbook of Trigonometry, , - sec, , 2 x, , 1, 5/3, '3, sin 0 = - or sin 0 = —, 2, 2, , =>, , 2, , • Ex. 44. Solve the equation 1 + 2 cosec x =, 2, , 0=H7C + (-l)n 6, , 2 x, , - sec —, 1 + 2 cosec x =--------- 2, , Sol. Here,, , 2 [2 + sin x] = - 1 + tan 2, , => General solution set is,, , £, 2, , 7t, , 7t, , 71, 2it, or 7t----- =, 3, 6, 3, . ., 7i 7t 2n, .*. Solutions are —, —, — and —., 6, 6 ’33*33, 6, , put n = 1,0 =Tt -, , 1 +tan2 —, , 2J, , Put tan — = t,, 2, , =, , 5n, , • Ex. 46. Solve the following system of equations., • 2, , sin x + cos y = 1, cos 2x - cos 2y = 1, , 2t, 2t, + 2 =-(l + f2), 1 + t2, 1 + t2, , Sol. Given,, , t3 + 2t2 + 3t + 2 = 0, , -(ii), , Put sin x = u and cos y = v in Eqs. (i) and (ii),, , (t + l)(t2 + f + 2) = 0, , u+v=l, , [vt2 + t+2*0], , t=-l, , c 1, , Tt, , •, , and, , u2+v2=2, , U, , k, , 1, , 1, , 1, , J, , Solving above equations; u = ~ and v = - => sin x = 2, 2, 2, , tan — = t = -1, 2, X, , (>), , 1, sin2 x + cos2 V = 2, , i.e., , 2t3 + 4fz +6t + 4 =0, , or, , sin x + cos y = 1, (1 - 2 sin2 x) -(2 cos2 y - 1) = 1, , and, , 4t2 + 4t + 4 + 2t (1 + t2) = 0, , x = mt +(- 1)" —, n e I and cos y = -, , 71, , „, , ., , , 71, , y = 2rmt ± —, m G I, 3, .'. The given equations have solutions,, , =>, , • Ex. 45. Find all values ofG lying between 0 and 271, satis­, fying the equations, r sin Q = 45, ...(i), , Sol. We have to solve for 0,, .*. We shall eliminate r from Eqs. (i) and (ii),, , 2, , 6, , Thus, x = 2n7t - —, n e I is the required solution., , r + 4 sin 0 =2 (45 + 1), , 1, , 71, , ., , — = mt-----,ne I, 2, 4, , 71, , 71, , 6, , 3, , x = mt +(- 1)" —, n el andy= 2rmt ± —, me I, , • Ex. 47. Find the coordinates of the points of intersection, •••(ii), , H, , 71, , of the curves y = cos x, y = sin 3x if----- < x < —., 2, 2, Sol. The point of intersection is given by, , From Eq. (i),, , 71, , V3, r =-----sin 0, , 45, , + 4 sin 0 = 2(V3 + 1), sin 0, 5/3+ 4 sin2 0 =2(V3+1)-sin 0, , .•.From Eq. (ii),, =>, , JT, it, , put n = 0,0 = — or —, 6, 3, 2 tan —, 2, , 1 + tan2 2, , 1 + tan2, , —, , sin x, , 2 x, , +2, , f, , IT 1, , ■0|0 =nn+ (-!)" - u 0|0 = nn + (-l)n 6I, I, 33, , 2 tan, , i.e., , 3, , 2, , sin x, , 2, , 0 = mt +(- l)n -, , or, , ,x, 1 + tan‘ —, 2, , i.e, , 7t, , sin 3x = cos x = sin -------X, 2, J, , it, 3x = mt + (- 1)" — - x, 2, , (i) Let n be even i.e. n = 2m, „, «, (n, 3x = 2rmt +-----x, 12, , www.jeebooks.in, 4 sin2 0-245 sin 0 - 2 sin 0 + 45 = 0, , =>, , 2 sin 0 (2 sin 0 -45) - 1 (2 sin 0 - 45) = 0, , (2 sin 0 - 1) (2 sin 0 - 45) = 0, , rmt it, x =— + —, 2, 8, , ...<0

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , (ii) Let n be odd i.e. n = 2m + 1, , =>, , 2A+B = —, 6, , 71, , 3x = (2m + 1)71 - ------X, 2, „, , „, , 71, , „, , „, , 71, , 2A = —, 6, , 7t, , 3x = 2rmt +— + x => 2x - 2rmt + —, 2, 2, , =>, , 157, , 12, , 4, , it, , x = rmt + —, 4, , [from Eq. (ii)J, , 3, , (ii), • Ex. 50. Find the number of values of 0 in the interval, , Now, as - — < x < —, 2, 2, , - —, — , satisfying the equation,, , It It, 37t, X = —, —,-------, , [From Eqs. (i) and (ii)], , 8 4, 8, Thus, point of intersection are, (it, TtA It, Tc'), —,cos — ,, I8, 8J 4, 4J, , I 2 2), , (1 - tan 0) (1 + tan 0) • sec2 0 + 2, , = 0. Also, find, , 71 71, —, — , satisfying the given equation., 3 3, , 06, , 371, 371, —, cos —, 8, 8, , tan2 0, , Sol. Given, (1 - tan 0) (1 + tan 0) sec2 0 +2*“5 99 =0, • Ex. 48. Find the range ofy such that the equation in x,, , =>, , (1 - tan2 0) (1 + tan2 0) + 2‘“2e, 2® =0, , y +cos x = sin x has a real solutions. Fory = i,find x such, thatO <x<2it., , 1-tan4 0+2‘“■'•=0, 20, , By observation, we have tan2 0 = 3., , 7t ), X------, , =>, , 4J, , -41 <y <41, , =>, , 0 = mt ±, , Tc'l, , 3/, Moreover there will be values of 0, satisfying, 3 < tan2 0 < 4, and satisfying the given equation as if f(x) = x2 - 2X - 1,, , ...(i), , Now, for y = 1, , then /(3+) f(4~) <0., , sin x - cos x =1, 11.1, 1, — sin x —7= cos x = —p, 2, 12, V2, =}, , 7, 1 + 2!t“2e = tan40, , =>, , ., 1, Sol. We have, y = sin x- cos x =41, sin x —7= cos x, ,V2, v2, , 1, , So the number of values of0 is 4., 71 It, , And 0, lying in the interval - —, — is ± —., 3, , 7t, , • II x----'* ---I sin, • I —, sin, 4J, 4, , • Ex. 51. Find the general solutions of the equation, x - — = nit + (- 1)" —, 4, 4, , =>, , (, , x = mt+(- 1)"- + 4 4, it, X = —,71, , 2, , 1, • Ex. 49. A triangle ABC is such that sin (2A + 8) = —. If A,, 2, B, C are in AP, then find the value of A, B and C., , x, x, A, f, x, 5, cos----- 2 sin x | sin x + 1 + sin----- 2 cos x cos x = 0., I, 4, J, , Sol. Here,, (, x, \, (, X, X 22 cos x | cos x = 30, cos---- 2 sin x sin x + 1 + sin----4, J, I 4, x, => cos — • sin x - 2 sin2 x + cos x + sin — • cos x - 2 cos2 x = 0, 4, 4, , • cos x + cos x - 2, , sinx- co, , (sin2 x + cos2 x) =0, , Sol. We have, sin(2A + B) = ~ = sin 71, 67, , sin, , 71, , 2 A +B = mt +(— l)n —, 6, Also, we have, A + B + C =it and2B = A +C, , X, , + cos x = 2, , 4, , Since, the greatest value of sin, , j and cos(x)is 1., , Therefore their sum is equal to 2 only if,, j = 1 and cos(x) = 1., sin, , www.jeebooks.in, =>, , 3B = 71 =s B = 3, From Eq. (i), for n = 1, , •••(ii), , — = 2mt + — and x = 2bt (n, k e I), 4, 2

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www.jeebooks.in, 158, , Textbook of Trigonometry, , Since, we have two choose those values of x which satisfy, both of these equations, we have, 8n7i, , ., , or, , cos2, , 271, , 2, , 2krt =------+ —, 5, 5, , =>k =, , 4n + 1, , -, where both k and n are integers. We write,, (n~l), , =>, , =>, , 5, , _, , 2, , 3, , sin2a< —, 4, (sina + l/2)(sina-l/2)<0, 1, ., 1, — < sin a < 2, 2, It ,, 7t, ,, , 71-1, , for------ = m, we have n = 1 + 5m and k = 1 + 4m (me I), 5, x = 2n + 877i7t, m e I., , .’. TTiTi---- < a < Turn- —, where me I, 6, 6, , • Ex. 52. Find all possible triplets (x, y, z) such that, , Clearly, x - 0 = 2mt ± cos, , 2, , where cos 0 =, , • Ex. 54. Find the solutions of the equation, , n, , (sinx+cosx)sin2x = a(sin3 x + cos3 x) Located between —, and 7t and for which values of ‘a’ does this equation have at, , => (sinx + cosx) sin2x - a |1 - -sin2x, (, 2, , =0, , •(»), , 37t, Now, sinx + cosx = 0 => tanx = -1 = tan —, 4, 371, X =---, , 4, , 7t, •/ — < X < 71, , 2, 71, , Hence, there is always at least one root lying in — and 7t for, 2, any value of the parameter a., Now,, 2sin2x — 2a + asin2x = 0, [from Eq. (i)], • «, 2a, sm2x =-----2+a, , ,, 2, cos(x- 0) =—, r, Since,, , ...(i), , This equation has real solution if,, _____ 2, =<1, 71 + 4 cos 2 a, , + 4 cos2 a>, , [sin2 x + cos2 x - sin x cos x], , where r2 = 1 + 4 cos2 a, =}r = 71 + 4 cos2 a, , cos2 a, , 2 cos a, , = or 0 = cos 1, fl +4 cos 2 a, , Sol. (sin x + cos x)sin2x = a(sin x + cos x), , Sol. Given, sin x +cos (a + x) + cos (a - x) = 2, 1 • sin x + 2 • cos a ■ cos x = 2, 1 = r sin 0,2 cos a = r cos 0, Let, r (cos x cos 0 + sin x • sin 0) = 2, , 2, , /, , 2 cos a, , 7t, , • Ex. 53. For every real number find all the real solutions, to equation sin x + cos (a + x) + cos (a - x) = 2., , cos(x - 0) =, , 71 + 4 cos, < 2 a,, , most one solution satisfying the condition — < x < 7t., , 2-2, , x = 4k, y = - 2k, z = 0, or, x = 2k,y = - k, z = 0, i.e. (2k, - k, 0) for any k e R., Hence, there are infinite number of triplets., , r cos (x - <}>) = 2,, , _____ 2, , =0, , _A_= y = z =k, 1-3, , [From Eq. (i)], , <7rj + 4', , x = 0 + 2nn ± cos, , => (2x +2y) +(2y + 4z) cos 20 +(z - x) - (z - x) cos 20 = 0, =>, (2x + 2y + z - x) + (2y + 4z - z + x) cos 20 = 0, =>, (x + 2y + z) + (x + 2y + 3z) cos 20 = 0, Which is zero for all values of 0, if, x + 2y + z = 0 and x + 2y + 3z = 0, x + 2y + z = 0, and, x + 2y +3z =0, , 6-2, , \, , cos2 a 7, , Sol. We have,, 1 - cos 20, , ,, , _____ 2, , (x + y) +(y + 2z)cos 26 +(z -x) sin2 9 = 0, for all B., , (x +y) + (y +2z) cos 20 +(z -x), , 3, , -cos a < — => 1 - cos a < 1 —, 4, 4, , 5, , k=n-, , 4, , -(ii), , 71, , — < x < it or it < 2x < 2it, 2, , -1 < sin2x < 0, Now, from Eqs. (ii) and (iii), we have, 2a, -1< — < 0, where a # - 2, 2+a, , .(iii), , www.jeebooks.in, or, , cos2 a or 4 cos2 a > 3, , 2a, 3 2a, -1< — and------ <0, 2+a, 2+a

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 0<, , 2+a, , • Ex. 55. Solve the equation, , + 1 and a(2 + a) < 0, , (tan x)e“2’ =(cotx)sinx., , Using number line rules, we get, ------------ H1, , /, ,2 _, , .CO J, , Sol. (tan x)', , -2/3, , a <-2, , or, , a>-~, 3, , • ••(A), , —-------, , and, , 0, , -2 < a < 0, , ■(B), , From (A) and (B), , -~<a<0, ...(iv), 3, (i.e. common to both (A) and (B)], Hence, for every value of ‘a’ satisfying the condition, 2, 2a, -- < a < 0 the equation, sin2x = ----- has the roots lying, , tanx J, .2, , (tan x)coC x+uni = 1, , 7t, , x = rm + —, 4, Case II When tan x * 1, 0 and cos2 x + sin x = 0, 71, , rm,rm + —, , or, , x, , and, , 1 - sin2 x + sin x = 0., , =>, , sin2 x - sinx - 1=0, , 4, , l±7T+4, , 75 + 1, , sin x =----2-------- =--------- ,, 2, 2, Here,, , > 1 which is not possible., , sinx =, , = sina, , '1-75, \, , 7t, , 2 +a, , 371, , <2x-2n<0, 2, , • „, 2a, sin2x =-----2+a, 2a, sin(2x - 2n) =-----2+a, , 2a, , 2+a, , 1 . . '2a', x = it + -sin, ,2 + a,, 2, , 2, , Thus, x = n7t + — or x = nn+ (-l)"a,, 4, f, rx, 1-V5, where a = sin, 2, , 2a, , — < 2x < 271, 2, , 2x - 2tc = sin, , imi, , x = n it + (-l)"a, where a = sin, , ., ., 2a, it 1 . ., sm(2x - it) =-------- ; x =-------- sin, 2+a, 2 2, , Since,, , ], , =>, , 0 < 2x - 7t < —, 2, 2a, sin(2x) =, 2+a, , Case II, , 1, , Now, here two case arises,, Case I When tan x = 1, the power (cos2 x + sin x) can take, any value., tanx = 1, , 7t < 2x < 27t, „, 3tt, 7t < 2x <--2, , Case I, , X, , =>, , between — and 7t., 2, Now, we have to find the solution of the equation, . _, 2a, 71, 2, sin2x =------ , where — < a < 0 and — < x < n, 2+a, 2, 3, , Clearly,, , 159, , • Ex. 56. Solve the equation, , acosx + cot x + 1 = cosec x., cos x, 1, Sol. acosx + ------ + 1 =------sin x------- sin x, asinxcosx + cosx + sinx = 1 (sinx * 0), sin x + cos x = 1 - asin x cos x, On squaring both sides, we get, 1 + 2sinxcosx = 1 + a2sin2xcos2x - 2asinx-cosx, a2sin2xcos2x - 2(a + l)sinxcosx = 0, sin x cos x[a2 sinx cos x - 2(a + l)] = 0, , sin2x[azsin2x - 4(a + 1)] = 0, , 3?r _, ., ., —, for a <— (-<», «>), 4, 7t 11 . . ( 2a ], -2, Thus, x =, for — < a < 0, —sin, k2 + a J, 3, 2 2, -2, ', 2a, ', 1 . ., for — < a < 0, it + -sin, 3, 2, +, , sin 2x = 0 for any value of a., 2x = nit, rm _, ,, ., x = — for a G (—°°, °°), 2, a2sin2x - 4(a + 1) = 0, , www.jeebooks.in, or

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www.jeebooks.in, 160, , Textbook of Trigonometry, , sin2x =, , 4(a + 1), a2, , Again, on subtracting the two given equations,, sinxcosy - cosxsiny = a2 - a, , v -1 < sin2x < 1 for all values of x., 4(a + l), -1 <, <1, a2, -a2 < 4(a + 1) < a2, , sin(x - y) = a2 - a, , As we know,, - 1 < sin(x - y) < 1, - 1 < a2 - a < 1, , -a1 < 4a + 4 or 4a + 4 < a2, , i.e., , -1 < a2 - a, , a2 + 4a + 4 > 0, , 1, a—, 2, , (a + 2)2 > 0, , (a - 2)2 - 8 > 0, , or, , and, , a2 - a + 1 > 0 and, , a2 - 4a - 4 £ 0, , or, , a2 - a < 1, a2 - a - 1 < 0, , a - 2 > 2V2, =>, , a - 2 < - 2V2, a < 2 - 24i, , or a>2 + 24i, , ae, , or, , ), , 2, , 1-V5 V5-1, 2, , ’, , 2, , sin3x = a sin x +(4 - 21 a |) sin2 x is a root of the equation, , Sol. We have,, , si n 3x + cos 2x = 1 + 2 si n x cos 2 x and any root of the latter, equation is a root of the former., , Sol. The first equation of the system can be written as;, 3 sin x - 4sin3 x = asin x + (4 - 21 a | )sin2 x, , sinx{4sin2x + (4 -2|a| )sinx +(a - 3)} = 0, , adding above equations,, sin xcos y + cosxsiny = a2 + a, sin(x + y) = a2 + a, , As we know,, , sin x(2sin x - 1) = 0, v Both equations have a common solution, therefore, , -1 < sin(x + y) < 1, -1 < a2 + a < 1, , •(ii), , sinx = 0, Also, second root of Eq. (ii) i.e., , -1 < a2 + a and a2 + a £ 1, , a;2 + a + 1 > 0 and a2 + a - 1 < 0, , sin x = -, satisfy Eq. (i)., 2, , 3^n, 1, + — £ 0 and I a + 2 I, 4, , 1-1.0, 4, , 1, true for all real 'a' and---- - < a + 2, 2, , (0, , The second equation is,, sin3x + cos2x = 1 + 2sinxcos2x, sin3x + cos2x = 1 + sin3x - sinx, cos2x = 1 - sinx, 1 - 2sinzx = 1 - sinx, , sinx cosy = a2 andcosx-siny = a, , 2J, , <, , • Ex. 58. Find all the numbers a ’for which any root of the, equation, , • Ex. 57. Find the values of ‘a’ which the system of equa­, tions sin x - cos y = a2 and sin y ■ cos x = a have a solution., , 1V, , (iv), , . a2 J, , for a e (-~, 2 - 2^2] u [2 + 2V2, «>), , =>, , 2, , 7, , <a<, , 2, , ---- ; for a E(-°o,00), 2, x=4(a +1)1, n it x (-Dn ., ------ b- -—-sin, 2, 2, 2, J, , i.e., , 2, , 1 + 45, , ' (l-45, , 4(a + 1)^, , rm (-1)", x = — + i—— sin, 2, 2, , 4, , From Eqs. (ii) and (iv) common solution is, , 4(a + l)>, a2 ), , 2x = rm + (-l)"sin, , =>, , <a<, , <, , a G (—°°, 2 — 2^/2] O [2 + 2V2,00), , or, , 1, a—, 2 1, , + — > 0 and, 4, , •>J55, 1 V5, => True for all real ‘a’ and------ < a — < —, 2, 2, 2, , which is true for all ‘a’ or (a - 2)2 > 8, , or, , (iii), , 2, , From Eq. (i), 4^| + (4 - 21 a I)-+ (a - 3) = 0, ', 2, 1 + 2 -1 a | + (a - 3) = 0, , —, 2, , www.jeebooks.in, (4~S + ? Ha<, \, , 2, , 7, , •••(ii), , I a I = a or a > 0 for sin x = 2, , .(iii)

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , Again from Eq. (i);, 4sin2 x + (4 - 21 a | )sin x + (a - 3) = 0, , For real x,, , Since,, , -1 < cos2xforallx., , -1 < cos2x < -,, 2, , (4 - 21 a | )2 - 4-4-(a - 3) > 0, , Using figure, we get 2rm + — < 2x < 2n7t + —, 3, 3, n, 5n, =>, rm + — <x<rm + —, 6, 6, , (2-a)2 -4(a-3)>0, =>, , 4- 4a + a2 -4a + 12 >0, , =>, , a2 -8a+ 16 >0, , 161, , (a - 4)2 > 0, , • Ex. 60. Solve the inequality,, olll A —, , ', , sinxcosx + -tanx>1, 2, , 2-4, , |o|-2± 7(o - 4)!, , =>, , or, =>, , Sol. Left hand side is defined for all x except,, , sinx = -—!-------- —------ —, 4, | a | -2+a-4, sinx =L, 4, |a|-2-a+ 4, smx = -—!-------------4, , Putting y = tan x, we get, , sinx = ——- or sin x = - [v | a | = a from Eq. (iii)], 4, 2, , a-3, , ., , l<a<5, Hence, a e [1,5]., , or, , (1+/), , 7, , 1 + y2 > 0 for all y., 2y + y(l + y2) - 2(1 + y2) > 0, yJ - 2y2 + 3y - 2 > 0, , or, , y2(y - 1) - y(y - 1) + 2fy - 1) > 0, , or, , (y - l)(y2 - y + 2) > 0, 1 i2, , where y2 - y + 2 = y —, 2, , (y - 1) >0or tanx > 1, From figure, we get, , Sol. Re-writing the inequality in the form,, 5(1 - cos2x) + 2(1 - cos22x) > 8cos2x, , 2cos22x + 13cos2x - 7 < 0, , Putting y - cos2x, we get, 4, n/3, , <4, , Y, , X = J x=|, , Hence, the solution of the inequality lies in the interval,, n, 7t, nn + —<x<rm + —,nel, 4, 2, 7t, n, x G n it + -, n tt + —, i.e., 4, 2, , COS5=4, , ^Tt/3, , y = tanx, >y=1, -------- >-X, , X-, , -sin2 x + sin2 x-cos2 x>cos2x., 4, , 7, , I + - > 0 for all y., , y, , • Ex. 59. Solve the inequality, 5, , or, , 2?, 1+/, , 1, , sm x =------ or sm x = 2, 2, For real x the values of ‘a’ will be suitable in the following, three cases (also a > 0)., a-3, ------ = 0 => a = 3, (i), 2, .... a-3 1, a=4, n ------ = 2, 2, a-3, <1 or |a -3| <2or-2 < a-3<2, (iii), 2, , =*, , i, , n, , x = rm + —, where net, 2, ,, 2tanx, Now, we have--------- — + tan x £ 2 {from given equation}, 1 + tan2 x, , x, , • Ex. 61. lf0<x< 2n, then solve the inequality,, 1, , COS— “ x, , 3, , 2, , 2 cosec, , www.jeebooks.in, 2y2 + 13y - 7 < 0 or (2y - l)(y + 7) < 0, , .1, 1, or y lies between -7 and - or -7 < cos2x < 2, 2, , V2, , Sol. The given inequality can be written as;, 2e“e'2l7(y-1)2 + i<2, , (i)

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www.jeebooks.in, 162, , Textbook of Trigonometry, , or, , Since, cosec2x > 1 for all real x., 2cose<:“i, , ., , >2, , On solving the inequality, we get, 1, y < — or y > 2., , ...(ii), , Also, (y - I)2 + 1 > 1, , -1)2 + 1 - 1, , =>, , From Eqs. (ii) and (iii),, 2coseA^y +, , ...(iii), , +j >2, , (y-2)(2y + 1) > 0, , But from condition y < -, we have y < --., 2, 2, , -(iv), , sin x < —1, 2, , or, , From Eqs. (i) and (iv), equality holds only when,, , On solving the inequality ( from graph we have), , - i)2 + i = i, , 2cosec2x = 2 and, , 1 •, -5n -7t /", 6 /, , cosec2x = 1, , or, , (y-l)2 + l = l, , and, , N, , sin x = ± 1 and y = 1, it 3n, ,, x — —, — andy = 1, 2 2, , or, , or, , —1, , 71 3ti, , 571, , Hence, the solution of the given inequality is x = —, — and, 2, y = l., , „, n, L, 5ti, 7t, it, tt o, Thus, x = 2n7t + —orxe 2nit----- ,2nit—, 6, 2, 6, , rr-l, , Ex., , 7-4sin2x<3|2sinx-l|, , Putting y = sin x,, , Now, consider the two cases, , (i) 2y - 1 > 0 or y > -, , then., , 7 - 4y2 < 3(2y - 1), [v for y > a, | y - a | = (y - a)], , or 4y2 +6y- 10>0 or, or, , 2y2+3y-5>0, , (y-l)(2y + 5)>0, , Solving this inequality, we get y < - - or y > 1., But from the condition y > -, we have y > 1., , Le., , sin x > L, , The inequality holds true only for x satisfying the, equation sin x = 1 that is, it, , ,, , x = 2mt H— when n e I., 2, 1, (ii) Let 2y - 1 < 0 or y < ', 2, then,, , 7 - 4y2 <-3(2y - 1), , 63. Solve, , | cos x - 2 sin 2x - cos 3x | = 1 - 2 sin x - cos 2x., , Sol. The given inequality can be written as;, 5 + 2(1 - 2sin2x) < 31 2sinx — 11, , 7 - 4y2 < 312y - 11, , 7t7t, , we get, 2nn------ < x < 2nit-----6, 66, , • Ex. 62. Solve the inequality,, , or, , k .1/2, , ...(i), , Sol. Here, LHS =|cosx - 2sin2x - cos3x|, = |(cos x - cos3x) - 2sin2x|, = |2sin2x-sinx - 2sin2x|, = |2sin2x(sinx - 1)|, and RHS = 1 - 2sin x - cos2x, = 2sin2 x - 2sinx = 2sin x(sin x - 1), , Thus, | cosx - 2sin2x - cos3x | = l-2sinx - cos2x, could be rewritten as,, 12sin2x (sin x -1)| = 2sin x(sin x - 1), where 1 - sin x > 0, for all real x, 12sin2x | (1 - sin x) = 2sin x(sin x - 1), (| 2sin2x | + 2sin x)(l - sin x) = 0, Either sin x = 1, or 41 sin x 11 cos x | + 2sin x = 0. So, two cases arises, Case I sin x = 1, „, , 71, , x = 2n 7t + —, n 6 I, 2, Case II 21 sin x 11 cos x | + sin x = 0, If sin x > 0, 2sin x | cos x | + sin x = 0, 1, or, COS X I = —, 2, Thus, no solution for, sin x > 0., Consider sin x < 0, =>, -2sinx | cosx | + sinx = 0, , (not possible), , www.jeebooks.in, [•.• for y < a, | y - a | = -(y - a)], , or, , 2y2 - 3y - 2 > 0, , sin x = 0 i.e. x = n it, n e I.

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www.jeebooks.in, 163, , Chap 02 Trigonometric Equations and Inequations, , i, , or, , =>, , i, , COS X, , 1, = -, , 2, , cos x = ± - and sin x < 0, 2, => Those values which lie in HI or IV quadrant, 71, k n + —, for k = (2m + 1), m G I, 3, X =, 7t, kn —, for k = (2m), m e I, ., 3, , 2mi + —,ne I, 2, , • Ex. 64. Prove that the equation 2 sin x = | x | + a has no, -------------------, , 3, , Sol. We have,, , k, 3^3-7t, , oo, , =>, , x|+, , (parallel to y2) intersects or touches the curve, , y! = sin x at least one point. In this case we must have, dy, 1, 7C, — = cos x = - (Le. the slope of the line) => X = —., dx, 2, 3, Hence, the solution exists if,, 3^3-71, 1 rr 1 _ . rt, ----- + -a<sm — => a<, 2 3 2, 3, 3, • Ex. 65. In &ABC, prove that, , 7, , > 0 and hence their arises three cases, 3, Case I When | x | > 2, we have, | x | + a > 2, whereas' 2sin x < 2, Hence the equation,, 2sin x = | x | + a, possesses no solution for | x | > 2, Case II When, -2 < x < 0, we have 0 < | x | < 2, =>, 2sin x < 0 and | x | + a > 0, The equation,, 2sin x = | x | + a has no solution., Case III When, 0 < x < 2, we have, In this case the given equation reduces to, Let, =>, , The equation sin x = - ( x | + - will have solution, if the line, 2, 2, y=, , Hence, x = (2n + l)n + -,ne I, 3, n, ft, 2nn-— ,ne I, 3, , solution fora e, , y,= sinx, , 2sin x = x + a, f(x) = 2sinx - x, f'(x) = 2cosx -1=0, 7t, x = — e (0,2) is a critical point., , A, B, (7, cosec—+ cosec — + cosec — > 6., 2, 2, 2, Sol. Since, , i.e., , ABC, — all are acute angles, we can use AM > GM., , ABC, cosec— + cosec — + cosec —, 2, 2, 2, 3, (ABC, > cosec— cosec — • cosec—, k, 2, 2, 2, , (0, , A, B, C 1, A (, xy, v, B . C, Consider, sin—• sin — -sin — = -sin — • I 2sin — • sin —, 2, 2, 2 2, 2, 2, 2, A( | B-C, 1 . A, = -sin— cos, - cos, 2, 2< I, 2, , 1 . Af |, , = -sm— cos, 2, 2V 1, , B-C, , 2, , 1 . Af, . , .., A, < -sm—| 1 -sin| —, 2, 2, 2, , it, f (x) =-2sinx < Ofor x = —, , 1/3, , B + Cp, , 2 J, , • 1, - sin, —, , I2JJ, , , as cos, , fB-C, 2, , <1, , - 1[ ■ A, . 2A |, < - | sm----- sin —, 2, 2, 2), , x = — is a point of maxima., , „ . 71 71, (Z(^))nux = 2sin------3 3, , 2ji, , ------- —, , 71, , z, , 3^3-71, , 3, , 2--- 3, 3^3-7t, a = 2sinx - x, cannot be greater, for the, 3, equation to have a solution. Hence, the result., , 1, a, Aliter We have, sinx = -| x | + -., 2, 2, , <1 1-, , "2l4, , AA, fl— sm, . —, |, \2, , 2, , <1, 8, , A, B, C, L, cosec, — • cosec—• cosec— > 8, 2, 2, From Eqs. (i) and (ii),, ABC, cosec — + cosec — + cosec —, 2, 2, 2 >(8)1/J, 3, A, B, C, cosec— + cosec— + cosec— > 6, =>, 2, 2, 2, , ...(ii), , www.jeebooks.in, Now, consider the graphs of yt = sin x and y, = -| x ].

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www.jeebooks.in, g Trigonometric Equations and Inequations Exercise 1:, Single Option Correct Type Questions, 1. The equation 2 sin — cos2 x - 2 sin — sin2 x, 2, 2, = cos2 x - sin2 x has a root for which the false, , 7. The general solution of 8 tan2 — = 1 + sec x is, , (a) x=2nrt±cos_1^-y-^, , (b) x=2rm± —, 6, , statement is, , (a) sin2x = l, , (b) cos x = 2, , (c) cos 2x = — 2, , (d) cos x = 1, , (c) x=2rm ± cos, , 8. The general solution of tan 0 + tan 40 + tan 70, , = tan 0 • tan 40 ■ tan 70, , 2. Let the smallest positive value of x for which the, X, , . . _, , = aq., , xsinc +ysin2c + zsin3c = sin 4c., Then, the roots of the equation., , x3 z o y + 2, z-x, r --t2, —1 +------- = 0(a, b, c * rm) are, 2, 4, 8, (a) sin a, sin b, sine, (b) cos a, cost, cose, (c) sin2a, sin2b, sin2c, (d) cos 2a, cos 26, cos2c, 5. The least positive value of x satisfying, , - 4 = 0is, , (b) 2, (d) 0, , 11. The number of real solution of the equation., sin(ex) = 5x +5‘x is, (a) 0, (b) 1, (c) 2, (d) Infinitely many, , 12. ABC is a triangle such thatsin(2A + B)=sin(C-A), = -sin(B+2C) = l/2. If A, B and C are in AP, then the, value of A, B and C are, (a) 45*, 60*, 75’, (b) 30*, 60*. 90’, (c) 20’, 60*, 100", (d) None of these, , 13. Let2sin2 x+3sinx-2>0and x2 -x-2<0(xis, = -is, 9, , (b)n/6, (d)57t/6, , 6. The maximum value of the expression, ■Jsin2 x + 2a2 - ^2a2 -1-cos2 x, where a and x are, , (c)V2, , - sin x, , (c) no real solution, (d) None of the above, , (a) >2, (c) 1, , x sin b + y sin 2b + z sin 3b = sin 4 b, , (b)2, (d)^, , (d) None of these, , 12, , 10. The number of the solution of the equation, COS (Tt^/x-4 ) • COS 71 Vx = 1 is, , 4. x sin a + y sin 2a+ z sin 3a = sin 4a, , real number, is, (a)l, , c e=—, , (a) x=0, (b) x=sin*'[log(2-V5)], , (b) 7, (d)9, , (a)n/3, (c)2n/3, , Hit, , 9. The solution of the equation e“nx -e, , 3. The number of solutions of the equation, 16(sin5 x + cos5 x) = 11 (sin x + cos x) in the interval, , 4 - sin2 2x - 4sin2 x, , 12, , 4, , Then, the sum of the digits in a is, (a) 15, (b) 17, (c) 16, (d) 18, , sin2 2x + 4 sin4 x - 4sin2 x cos2 x, , (b) e=—, , (a)0=—, , X, , function f(x) = sin — + sin —, (x e R) achieves its, 3, 11, maximum value be x0. Express x0 in degree i.e. x0, , [0,2n] is, (a) 6, (c) 8, , (d) None of these, , 3>, , measured in radian). Then *x’ lies in the internal., (n 5n, (b)[-1.^, (a) 6’T, \, , (c)(-1.2), , 6 ., , (d), , 14. The number of points of intersection of the two curves, y = 2sinxandy = 5x2 + 2x+3is, (a) 0, (c) 2, , (b) 1, (d) oo, , www.jeebooks.in

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www.jeebooks.in, 166, , Textbook of Trigonometry, , 15. The number of all possible triplets (a,, a2,a3) such that, , x2 + 4 + 3 sin( ax + b) - 2x=0 has atleast one solution,, , a, +a2 cos2x + a3sin2 x = 0for all xis, , then the value of (a + b) can be:, , (b) 1, (d) Infinite, , (a)0, (c) 3, , / A 771, , 16. The equation sin4 x-(fc + 2)sin2 x-(k + 3) = 0possesses, a solution if, (a) k>-3, (c) -3<k<-2, », , • ., , 25. If a, bE [0,7t ] and the equation, , (b)fc<-2, (d) k is any (+ ve) value, , (a) —, 2, 9k, , 3n, , •, , (c)T, , (b) —, 2, (d) None of these, , 26. The value of a for which the equation, 4cosec 2 (k (a + x)) + a2 - 4a = 0 has a real solution, is, , It 71, , i, , If' In, UlUlltiVCU, ,----- , the equation log linO(COs20) = 2haS, 17., interval -------—, . 2 2, (b) a unique solution, (a) no solution, (d) infinitely many solution, (c) two solution, , i-i, , (a)n-l, (c)n, , (b) 0, (d)n + l, (b)(- «,2], (d) None of these, , 20. The most general values of ‘x’ for which, sin x + cos x = min [1, a2 - 4a + 6] are given by, oeR, , (b)2nn + —, 2, , (a) 2nn, (c) rm +(-!)"--4, , 4, , +sec2 2y+ cos4 y are, , (c) x=l, y =2nn, , K, , (b) x=l, y=2rm + —, 2, (d) None of these, , 22. If max {5sin0 +3sin(0 - a)} = 7, then the set of possible, value of a is 0 g R, f, k, (a) < x:x = 2nrt±—,ne/, 3, , . n 2n, (c), [_ 3 3, , 29. The number of solutions of 'he equation, 5sec0-13 = 12tan0 in[0,27t]is, (a) 2, (b) 1, (c)4, (d)0, 30. The number of solution of equation, x3 +x2 +4x + 2sin x = 0in0<x<2n is, , (d) None of these, , 21. Value of‘x’ and 'y' satisfying the equation, sin7 y = |xs -x2 -9x +9|+|.xJ -4x-x2 +4, , (a) x=l,y=nx, , (b) A rational number, (d) None of these, , In, 28. Let n be positive integer such that sin — + cos — = 2, 2n, 2n, Then, (a)6<n<8, (b)4<n<8, (c)6<n<8, (d)4<n<8, , 19. If0<x<7t/2andsinn x + cos" x>l, then, , (a)ne[2,«>), (c)ne[-1,1], , (b)a=2, (d) None of these, , 27. If the equation 2 cos x + cos 2Xx = 3 has only one, , solution, then X is, (a) 1, (c) An irrational number, , n, , 18. If Xcos©, =n, then ^sin0( is, i-i, , (a)a=l, (c)a=10, , (a) Zero, (c) Two, , (b) One, (d) Four, , / 7t, 1, I 71, 1, 7t, 31. If tan —sin0 =cot —cos0 [ then sin 0 + cos 0 is, 2, <2, equal to, (a) 0, (b) 1, (c) -1, (d) 1 or -1, , 32. The equation sin x + sin y+sin z = - 3 for 0 < x < 2n,, , I, 2n, (b) ■! x:x=2m±—,nel, !, , 3, , (d) None of these, , 23. The number of integral values of ‘n’ so that, sin x(sin x + cos x) = n has atleast one solution, is, (a) 2, (b) 1, (c) 3, (d) zero, 24. Total number of solution of sin{x} = cos{x}, where {•}, denotes the fractional part, in [0,2n] is equal to, (a) 5, (b) 6, (c) 7, (d) None of these, , 0<y <2k,0<z<2k has, (a) one solution, (c) four sets of solutions, , (b) two sets of solutions, (d) no solution, , 33. If x = nn + (-l)na,nGl and x = n7t+(-l)"P are the roots, , of 4cos x-3secx = tanx, then 4(sina +sinP)is, (a)-l, (b)l, (c) 2, (d) None of these, 34. If tan mQ = tan n0 and general value of 0 are in AP, then, , common difference is, (a) —, m-n, , (b) —, m+n, , t \, , (d) None of these, , n, , www.jeebooks.in, (c)------, , m-n

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 35. Ifsin3a = 4sinasin(x+a)sin(x-a), then, , 44. Total number of solutions of cos x = A/l-sin2x in, , [0,2it], is equal to, (a) 2, (c) 5, , (a) x=mt ± —,nel, 3, , It, , (b) x=nn±~,nel, 6, , 167, , (b) 3, (d) None of these, , 45. If the equation cos3x cos3 x+sin3xsin3 x = 0, then x is, , (c) x=mt±—,nel, 2, (d) None of the above, , equal to, , (a)(2n+l)^, , 36. X cos x -3 sin x = X +1 is solvable only, if, , (a)Xe[0,5], (c)Xe(-«(4], , (b)Xe[4,5], (d) None of these, , 37. cos2x-3cosx + l = --------------- ---------------- holds, if, (cot 2x - cot x) sin( x - 7t), (a) cosx=0, , (b) cosx=l, , (c) cosx=2, , (d) for no value of x, , 38. Ifsecxcos5x = -1 andO<x<—, then xis equal to, 4, / \71, , W3, , l \n, , (d) None of these, , 7, , (b)(2»-l)4, , 4, , mt, , (d) None of these, , c7, , 46. Total number of solutions of sin x = — is equal to, 10, , (a) 4, (c) 7, , ., , (b) 6, (d) None of these, , 47. The number of all possible 5-tuples (ava2,ai,ai ,a5) such, , that a, +a2 sinx + a, cosx + a4 sin2x + a5 cos2x=0, holds for all x is, (a) zero, (b) 1, (c) 2, (d) infinite, 48. xt and x2 are two positive values of x for which, , 2 cos x, |cos x| and 3sin2x-2areinGP. The minimum, , 39. Ifsin'“ 0 - cos100 0 = 1, then 0 is, Tt, , value of|xI -x2| is equal to, 71, , (a)2nn + — ,nel, 3, , (b) mt + —,nel, 2, , Tt, , 71, , (c)nn + -,nel, , (d)2mt--,nel, , 3, (c)2cos, , 40. IfT3sinx-cosx = min{2,e2,7t,a2 -4a +7}, then, 49. If cos x3, , n it, (c) x=n7t+(-l)n —+—,neJ (d) x=nn+(-l)"+ ------ ,nel, 4 6, 4 3, rr, , It, , 41. The number of solutions of the equation, cos 4x + 6=7 cos2x, when xe[315°,317°]is, (a)0, (b)l, (c) 2, (d) 4, 42. The number of solutions of cot(57tsin0) = tan(57tcos0),, , V0e(o,2n)is, (b) 14, (d) 28, , 43. If exp [(sin2 x + sin4 x+sin6 x + ...««) In 2] satisfies the, equation y2 -9y+8 = 0, then the value of, , cosx, , n _, 7t ., ,0<x<—, is, 2 J, cosx + sinx, (a) 73 + 1, , (d) cos, , 3, , cot P sin x, , 2, , 2n, (b) x=2nit +—,nel, , (a) x=2rm,nel, , (a) 7, (c) 21, , 3, 2, , 73-1, , 3, , 73, x, = —, then the value of tan — is, 2, 2, , (a) tan-tan 15°, 2, (b) tan |, (c) tan 15°, (d) None of the above, 50. The expression n sin2 0 +2n cos(0 +a), , sin a sin 0 + cos 2 (a + 0) is independent of ‘0 the value, of n is, (a) 1, (b) 2, (c)3, (d) 4, , 51. The value of the determinant, a2, 1, a, cos(n-l)x, , cos nx, , cos(n + l)x is zero if, , sin(n - l)x, , sinnx, , sin(n + l)x, , (a) x=mt, , (b), , 2, (d) None of these, , 2, , (c) x=(2n+l)n/2, , (b) x=mt/2, 1 + a2, (d)x= ------ nel, 2a, , www.jeebooks.in, (c)73-1

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www.jeebooks.in, 168, , 52., , Textbook of Trigonometry, , 57. If cot (a + P) = 0, then sin (a + 2P) =, , < 0 if a lies in, cos 2a, (a) (1371/48,147t/48), (c) (187t/48,237t/48), sin20, , 53. Iff(x)= cos2 0, x, , (b) (1471/48,187t / 48), (d) any of these intervals, , cos220, x, sin20, , x, , sin2 0 0 G (0,7t / 2\ then roots, , cos2 0, , (a) sin a, (c) sinP, , (b)cosa, (d) cos2p, , 58. If cot 0 + cot I — + 0 I = 2, then the general value of 0 is, 7t, , 71, , (b) 2nn ± —, 3, , (a) 2rm ± —, 6, , 71, , 7t, , of /(x) = 0are, (a) 1/2,—1, (c)-1/2,1,0, , (d) rm ± —, , (c) rm ± —, (b) 1/2,—1,0, (d) -172,-1,0, , 54. The equation sin x + sin y + sin z = - 3 for 0 < x < 2n,, 0<y<27t,0<z<27t, has, (a) One solution, (b) Two sets of solutions, (c) Four sets of solutions (d) No solutions, , f, , 1 A, , 59. If cos 20 = (Vi + 1) cos 0 —— I then the value of 0 is, I, yl2j, , (a) 2rm + —, , (b) 2n7t ± —, 4, , (c)2n7t-^, , (d) None of these, , 55. Ifsecxcos5x + l=0, whereO<x<27t, then x =, . , It 71, , z, , 7t, , (c)7, , (d) None of these, , (a)?7, , 1+sinx, , > -, then sin x lies in, 3, , -1 1, , J -1, 'A I 2 J 1.2 ., , 56. If | k | = 5 and 0? < 0 < 360°, then the number of different, solutions of 3 cos 0 + 4 sin 0 = k is, (a) Zero, (b) Two, (c) One, (d) Infinite, , l 2 2J, - J 11, L 2 2J, (d) None of the above, , g Trigonometric Equations and Inequations Exercise 2:, w More than One Option Correct Type Questions, 61. The value of 7’ which satisfies (t — [| sin x |])! = 3! 5! 7!, , is/are............ where [.] is GIF, (a) 9, (b) 10, (c)H, (d) 12, 62. Let f(x) = cos (a, + x) + - cos(a2 + x) +, cos(a3 + x), 2, 2, 1, +... +------ cos(a„ + x), 2"-1, , where a,,a2...... an G R. If f(xx) = f(x2) = 0, then, | x2 — x, j may be equal to, (a) 7t, (b) 2tt, (c)37t, , chords AP, AQ and AR are in GP where A is (1,0), then, [Given a, p, y G (0,2n)]., ... a + y, a-Y, a-Y . . P3, (a) sm------- cos----- - £ sin 4, 4, 2, (b) sin^^^cos^ a-y, <sin—, 4, 2, / \ ■ a • Y, ■ P, (c) sin—sin- > sin2, 2, 2, , .x . a . Y, • P, (d) sin—sm- <sin2, 2, 2, , 64. Let x, y, z be elements from interval [0, 2jc] satisfying the, , inequality (4 + sin 4x) (2 + cot2 y)(l + sin4 z) < 12sin2 z,, , 7T, , (d)2, , 63. Let a, P, y be parametric angles of 3 points P, Q and R, , respectively lying on x2 + y2 = 1. If the lengths of, , then, (a) the number of ordered pairs (x, y) is 5, (b) the number of ordered pairs (y, z) is 8, (c) the number of ordered pairs (z, x) is 8, (d) the number of pairs (y, z) such that z = y is 2, , www.jeebooks.in

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 65. The number of integral values of a for which the system, of linear equations x sin 0 - 2y cos 0 - az = 0,, x + 2y + z = 0,-x + y + z = 0 may have non- trivial, solutions, then, (a) at a = 2 the given system will have finite solutions for, 6eR, (b) number of possible integral values of a is 3, (c) for a = 1 the system will have infinite solutions, (d) for a = 3 the system will have unique solution, , 66. The equation, 2sin3 0 +(2X -3)sin2 0 -(3X + 2)sin0 -2X = 0has, , exactly three roots in (0,2n), then X can be equal to, (a)0, (b)|, , 67. Ifx + y = 27t/3andsinx/siny = 2, then the, (a) number of values of xe[0,47t] are 4, (b) number of values of xe[0,47t] are 2, (c) number of values of y e [ 0,4n ] are 2, (d) number of values of y e[0,4n] are 8, , 68. If 0 < x < 2n and |cos x| < sin x, then, (a) the set of all values of x is —,—, 4 4, , (b) the number of solutions that are integral multiple of — is, 4, four, (c) the sum of the largest and the smallest solution is 7t, zjv, i, ii, t, /*, •, 7t 7t |, i 7t, n 3n, (d) the set of all values of x is x e —,— lu —,’ 4, .4 2/ <2, 69. If x and y are positive acute angles such that (x + y) and, (x -y) satisfy the equation tan2 0 - 4 tan0 +1 = 0, then, 7C, , n, ^y=4, (d)x=^, , na, , (a) x=6, (c)y=7, 6, , 4, , 70. Ifx + y =—and sin x = 2siny, then, 7t, , / \, , 72. If[sinx] + [V2cosx] = -3, xG [0,271 ], (where, [•] denotes, the greatest integer function), then, (, , / X, , 571, , zux, I 7jt, (b) xe it,—, \ 6, , (a)x6 7t,—, \ 4, , .., , r 5ti, , (d) None of these, , (c)xe 7t,—, 4, , 73. If a G [- 271,271 ] and cos — + sin — = Ji (cos 36°- sin 18°), 2, 2, then a values of a is, , (a)T, , (d)-£, , (C)-^, 6, , 6, , 74. The number of values of a in the interval [- 7t,0], , (d)-l, , (c)l, , 169, , (a) x=n7t +—,nEl, 2, , satisfying sin a + j cos 2x dx = 0, then, , (a) a=0, (b) a=0,-7t,-3, , ., It 571, , c a=-,—, 6 6, , (d) None of the above, , 75. The solutions of 0 G [0,271 ] satisfying the equation, , log^j tan0(^log„9 3+log3-73 ) = -1, then, , (a) 0 = 7, , (b)S^, , 6, / M, , 3, 471, , 3, , (d)>2, , (c) has sum —, , 76. If a and 0 are the solutions of a cos 0 + 5 sin 0 = c, then, o 2bc, (a) sina+sinp=—---- a +b, , (b) sinasinB=^~-r, a‘+b, (c) sina+sinP=-r~, b‘+c, (d) sina+sinP=, , c, , a^b2, , 77. The solution of the equation sin 2x + sin 4 x = 2 sin 3x is, , rm, , (b) y=--rm,nel, 6, , (c) Both (a) and (b), (d) None of the above, , 71. The number of solutions of the equations, , (b) x=rm, (c) x=2rm, (d) None of the above, , 78. The general solution of 4 sin4 x + cos4 x = lis, , y = - [sin 0 + [sin 0 + [sin 0 ] ] ] and [y + [y]] = 2 cos 0, 3, , [where, [•] denotes the greatest integer function] is /are, (a)0, (b)l, (c) 2, (d) infinite, , (b) rm, , (a)(2n+i)y, , www.jeebooks.in, (c) rm ± sin, , 4, , (d) None of these

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www.jeebooks.in, 170, , Textbook of Trigonometry, , 71, 79. The values of x, 0< x < — whcih satisfy the equation, 2, 81“*’* + 81"” X = 30 are, (b) —, , (a)I, , <4, , (c)7, , I ft I, 80. All values of x e p, — such that, , V. 2 J, , sin x, , cos x, , are, , "i, , (a)n, , nn, , 7t, 84. The value of 0, lying between 0 =0 and 0 = — and, 2, satisfying the equation., 1 + cos20, sin20, 4 sin 40, , cos2 0, , 1 + sin2 0, , 4 sin 40, , cos2 0, , sin2 0, , 1 + 4 sin 40, , (a)7r, 5n, , a2, , sin2 x+a2 -2, , 1-tan2 x, , cos 2x, , has a solution if, , 82. The equation 4sin(x + 7t/3)cos(x-7t/6) = a 2, + 73 sin 2x - cos 2x has a solution if the value of, , (a) -2, (c) 2, , (d) None of these, , (c)7, , 85. If [x] denotes the greatest integer less than or equal to x, , (b) a>l, (d) a is any real number, , (a) a<-l, (c)a=l/2, , =0, is, , f J171, , (C)1T, 81., , 83. Which of the following is/are correct., (a) (tan x)™’" x) >(cot x)ln (,in x), VxG(0,7t / 4), (b) 4'Inroscc,<5lnc^cr,Vxe(0,7t/2), (c) (l/2)l,n(“5r)<(l/3),n(co*x), VxG(0,7t/2), (d) 21‘ n(tanx)>2In(unx)Vxe(0(K/2), , (b) 0, (d)a,ae]-2,2[, , then the equation sin x = [1 + sin x] + [1 - cos x] has no, solution in, n 7t, (b) ^7:, (a), 2’2, 2, 371', (d)/?, (c) n,—, 2, , g Trigonometric Equations and Inequations Exercise 3:, Passage Based Questions, Passage I, (Q. Nos. 86 to 88), , If number of solutions and sum of solutions of the equation, 3sin2 x- 7sin x+ 2= Qxe [0,27t]are respectively N and S, and fn (0) = sin" 0 + cos'10. On the basis of above information,, answer the following questions., 86. Value of N is, (a) 1, (b) 2, , 90. If Nf is number of integers when a = 2 and a = 2 and N2, , is number of integers when a = 1 and a = 3, then the, minimum value of(Nt sec20 + N2 cosec20), (a) 10 + 4 76 (b) 10 + 76, , 87. Value of S is, 571, , 7, , 89. The difference of largest and smallest integral value of N, satisfying a = 3 and a = 5, is, (a) 499, (b) 500, (c) 501, (d) 502, , (b)T, 6, , (c)3, , (d)4, , (c) 271, , (d)n, , 88. If a is solution of equation 3 sin2 x - 7 sin x + 2 = 0,, x e [0,2n], then the value of f4 (a) is., (a)^, (b)^, (c)~, (d)0, , (c) 10, , (d) 100, , Passage III, (Q. Nos. 91 to 93), If an angle and a side of a right angle triangle is known, then, rest of the sides and angles can be found as follows, , In MBC (Figure 1), if Z5 = 9(F, ZC = 0 and BC = x, then, AB = xtan0 and JC = xsec9., A, , Passage II, (Q. Nos. 89 to 90), , www.jeebooks.in, Let log a N = a + 0 where a is integer and P = [0,1). Then,, , On the basis of above information, answer the following, questions., , B, , C

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , Now, consider an isosceles triangle PQR (Figure 2 ),, , 171, , 95. sin a +sin0+siny can be equal to, . . 14-372, , (a)------ 7—, , (b)5/6, , (c), , (d), , 672, 3 + 472, 6, , 1 + 72, 2, , 96. sin (a - 0) is equal to, 0, , ■R, , N, , (b)0, , (a) 1, . ,1-276, , where PQ = PR and 20N = -75, , (d)A±^, , (c)—, , 6, , On the basis of above information answer the following, , Passage V, , 91. The angle of triangle PQR are, (a) 150*, 15’ 15*, (c) 120’, 30’, 30’, , (Q. Nos. 97 to 99), , (b) 60*. 60*, 60’, (d) 75’, 52.5’, 52.5’, , Consider the equations, 5sin2 x+3sinxcosx-3cos2 x=2, , 92. Area of circumcircle of quadrilateral PLOM is, , (b)-, , (a)n, , 3n, , (c)—, , ., , 4, , sin2x-cos2r=2-sin 2r, , (d)3n, , 4, , ...(ii), , 97. If a is a root of (i) and 0 is a root of (ii), then, , (a) tan 15°, , (b) -J3 tan 15°, , tan a + tan 0 can be equal to, (a) 1 + 769/6, (b)-1-769/6, , (c) cot 15°, , (d) 73 cot 15°, , (c), , 93. Length of the side QR is, , ... 3-769, (d), , -3 + 769, , Passage IV, , ., , •••(*), , 98. If tan a, tan 0 satisfy (i) and cos y, cos 8 satisfy (ii), then, , (Q. Nos. 94 to 96), , tan a • tan 0 + cos y + cos 8 can be equal to, , a is a root of equation (2sinx-cosx)(l + cosx)=sin2 x, 0 is a, root of the equation 3cos 2x -1 Ocos x+3 = 0 and y is a root of, the equation 1 — sin 2x=cosx-sinx: O<a,0,y < n/2, , 5, , ..5 2, (c)-—, 3 713, , 94. cos a + cos 0 + cos y can be equal to, , 2, , (b)-; + ~r=, 3 713, 5, 2, 3 713, , (a)-l, , 99. The number of solutions common to (i) and (ii) is, (a)0, (b) 1, (c) finite, (d) infinite, , ,.376+272+6, (a)-^—, , ,LX 373-8, <b), (b) —, , (c)^±?, , (d) None of these, , g Trigonometric Equations and Inequations Exercise 4:, Single Integer Answer Type Questions, 100. Let At be the area of triangle APk B which is inscribed in, , a circle of radius 2 units. If AB diameter of circle,, /.ABPk = — and £ At =4 cot —, then - is equal to, 2n, t o| 32, 2, , 101. If the sum of the roots of the equation, cos 4x + 6 = 7 cos2x in the interval [0,314] is kit, k e R, , (where a, b, c, > 1) have a common root and then 2nd, equation has equal roots, then number of possible value, of 0 in (0, it) is, , 103. Number of ordered pairs (x, y) which satisfies the, relation, , x4 +1, , = sin2 y. cos2 y, where y e [0,2n],, , 8x2, , 104. The number of solutions for,, , Find (k - 4948)., , . f, , 102. If equation x2 tan2 0 -(2tan0)x + 1 = 0and, , r, , i, , Ax + z, , 1—.1 = 0, , nA, , (, , 3it, 3nA, , sin x — I - cos | x + — | = 1, I, 4, k, 4j, ■ in(0,2n), is, 2 cos 7x, >2CO*21, cos3+sin3, , www.jeebooks.in, ____ 1, , ■ x2 +, , ? + logfc ac), , <1 + logc, , U + logabc, , J

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www.jeebooks.in, 172, , Textbook of Trigonometry, , 105. If cos A sin I A - — is maximum, when the values of A is, , I, , 6J, , 108. If a be the smallest positive root of the equation, ^/sin^-x) = -Jcos x, then the approximate integral, , 71, , value of a must be, , equal to —, then the value of X is, , 109. If x and y are the solutions of the equation, 106. Let p.qeN and q>p, the number of solutions of the, equation q Jsin 0| = p |cos 0| in the interval [0, 2ti] is, , 107. If 01,0 2,0 3 are three values lying in [0,2n] for which, tan 0 = X, then, 0., 0,, 0,, 03, 0., 03 ., tan — • tan — + tan — • tan — + tan — • tan — is equal, 3, 3, 3, 3, 3, 3, to, , 12sinx + 5cosx = 2y2 -8y + 21, the value ofl2cot(^, must be, , 110. If tan(7t cos0) = cot (7tsin 0), then cos2 (0 — 7t / 4) is, equal to, , 111., , If 3 sin x + 4 cos x = 5, then the value of, 90 tan2 (x/2)-60tan(x/2) + 10is equal to, 1, , g Trigonometric Equations and Inequations Exercise 5:, Statement I and II Type Questions, ■ This section contains 6 questions. Each question contains, Statement I (Assertion) and Statement II (Reason)., Each question has 4 choices (a), (b), (c) and (d) out of, which only one is correct. The choices are, (a) Statement I is true, Statement II is true; Statement II, is a correct explanation for Statement I., (b) Statement I is true, Statement II is true, Statement II, is not a correct explanation for Statement II., (c) Statement I is true, Statement II is false., (d) Statement I is false, Statement II is true., 112. Statement I sin x = a, where -1 < a < 0, then for, x G [0, n7t] has 2(n -1) solutions V neN., Statement II sin x takes value a exactly two times, when we take one complete rotation covering all the, quadrants starting from x =0., , 113. Statement I The number of solutions of the equation, |sin x| = |x| is only one., , x + (sin a )y + (cos a )z = 0, x+(cosa)y+(sina)z=0, - x + (sin a)y - (cos a )z = 0, has a non trivial solution for only one value of a lying, between 0 and 7t., sin x cos x COS X, , Statement II cos x, , sin x, , cos x = 0, , COS X, , COS X, , sinx, , has no solution in the interval -7t/4<x<7t/4., 116. Let0G(7t/4,7t/2), then, , Statement I (cos0)unO <(cos0)C0*9 <(sin0)co’9, Statement II The equation e"”9 -er*1"6 =4 has a, unique solution., 117. Statement I If, • exp {(sin2 x+sin4' x+sin4 x +... inf) log, 2} satisfying the, , Statement II |sinx|>0VxG/?., 114. Statement I If 2sin2x-cos2x = l, x^(2n + 1)71/2, n is, the integer, then sin 2x + cos 2x is equal to 1 / 5., , Statement II sin 2x + cos 2x =, , 115. Statement I The system of linear equations, , COS X, , equation x2 -9x + 8=0, then the value of-------------- is, cosx + sinx, , l + 2tanx-tan2 x, , V3-1, , 1 + tan2 x, , 2, , •(0<x<7t/2), , Statement II sin2 x + sin4 x + sin4 x + ...inf = sec2 x, , www.jeebooks.in

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 173, , g Trigonometric Equations and Inequations Exercise 6:, Matching Type Questions, , ■ Math the statement of Column I with the value of, Column IL, , 118., , Column I, , (A), , If a, P are the solutions of, , a-p=n, , (P), , -V3, the solutions of cosx =------ in, 2, [0,2k], then, , If a, p are the solutions of, cotx= -/3 in [0,2n ]and a, y, , (q), , p-y=n, , Column II, (p) 0=3n/8, , (A), , 2sin0|cos0|=--^=, , (B), (C), (D), , 2cos20cos40 + 2cos320-l=O, (q), 8cos20sin0-4cos:0-2sin0+l=O (r), sin 40= ±1, (s), , If a, P are the solutions of, , 0=7n/8, 0=2n/3, 0=n/6, , (a) A (p, q); B ->(p, q, r); C ->(r, s); D -> (p, q), (b) A -»(r, s); B -> (q, r); C -> (r); D ->(p, s), (c) A —»(p); B —> (q); C —»(r); D —> (s), (d) A -»(s); B -> (q, r); C -»(r, s); D —> (p, q), rr r, sin0 + sin30 + sin 50 + ... + sin((2n -1)0), 120. Iffn(Q) =------------------------------------------------------ sin0 + cos0 + cos50 + ... + cos((2n -1)0), , are the solutions of cosec x=-2, in[0,2rt], then, (C), , Column I, , Column II, , sinx=~ in[0,2n]and a, yare, , (B), , 119., , Column I, , a-y=n, , (r), , (A), , sinx=~ in [0,2n ]and a, y are, the solutions of tanx=-4= in, , (B), , V3, [0,2tc ], then, (s), , a+P=3n, , (0, , P+y=2n, , (a) A —> (q, s); B —»(p, t); C -> (r, s, t), (b) A-> (q); B -> (t); C-> (r), (c) A —> (r, t); B —> (t); C -> (p, q), (d) A -> (p, q); B -> (q, r) C -> (r, s, t), , (C), (D), , Column II, , 5-1, , (P), , ■, , 4^, , (q), , (nA, , (16 ), , '•ET-, , (r), , V2 + 1, , (s), , 2+V3, , (0, , 1, , (a) A—>(p); B ->(q); C-> (r); D-»(s,t), (b) A-»(t); B—>(p); C—>(r); D-»(q), (c) A—»(q); B —> (r); C-> (s); D —> (t), (d) A—»(r,t); B->(s); C -> (p); D -> (q), , Trigonometric Equations and Inequations Exercise 7:, Subjective Type Questions, 121. Find the number of solutions of the equations;, , (i) | cot x | = cot x +, , 1, , when x G [0, 2n], , sin x, , (ii) sin3 xcosx + sin2 x.cos2 x + sinx.cos3 x = l,, , when x G [0, 2tt ], (iii) 2“*' = | sin x |, when x G [—271, 2n], , (iv) | cos x | = [x], (where [.] denotes the greatest, integer function)., , 122. Find all value of a for which the equation, sin4 x + cos4 x + sin2x + a = 0 is valid. Also, find the, general solution of the equation., , 123. If32tans 0 = 2cos2 a -3cosa and3cos28 = 1, then find, the general value of a., i, , 124. Solve for x and y, , 4«nx + 3co,y =11, i, , 5.16“nx -23c”y =2, , www.jeebooks.in, (v) x + 2 tan x - —, when x G [0,2n].

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www.jeebooks.in, 174, , Textbook of Trigonometry, , 125. Find all numbers x, y that satisfy the equation, ■, , 2, , ,, , 2, , 1, , sin x +-------sin2 x, , 1, = 12 +-siny., 2, , 1, , + cos2 X +, k, cos, , 133. Find all the values of 'a' (a * 0) for which the equation, , f (t2 - 8t + 13)dt = xsin — has a solution. Find the, Jo, x, solution., , 126. Find all the solutions of x, y in the equation, , 4 3y4x - x2 sin 2, x, , x+y, , + 2cos(x + y), , 2, , 134. Find all values between 0 and it which satisfies the, equation, 17, , 2„, , . sin8 x + cos 8 x = — cos 2x, 16, , = 13 + 4 cos2(x + y), 127. Solve for x and y, l-2x-x2 = tan2(x + y) + cot2(x + y)., , 135. Find all number pairs x, y that satisfy the equation, , tan4 x + tan4 y + 2cot2 x.cot2 y = 3 + sin2(x + y)., 128. Solve the system of equations, 136. Determine all values of‘a’ for which the equation, cos4 x - (a + 2) cos2 x - (a + 3) = 0, possesses solution., , tan2 x + cot2 x = 2cos2 y, cos2 y + sin2 z = 1, , Find the solutions., , 129. Find all pairs of x, y that satisfy the equation, , equation, , /, x, 3, cos x + cos y + cos(x + y) = —, 2, , /g\, , 130. Solve the equation cotl — I — cosecc, , (■Ji sin x + cos x), 0, , = cot 0., , 2, 3 • „, 131. Find the general solution of 1 + sin3 x + cos3 x = -sin, 2x., 2, , 132. Solve log(sinx) 21og(jinJ, , 137. For x E (-it, it) find the value of x for which the given, , a = -1 stating any condition on ‘, , d that may be required for the existence of the solution., , = 4 is satisfied., , 138. Show that the equation, sec 0 + cosec 0 = c has two roots between 0 and 2it, if, c2 < 8 and four root if c2 > &, , 139. Solve the equation for x and y,, | sin x + cos x |**nx-1/4 = 1 + | siny | and, , cos2 y = 1 +sin2 y., , □ Trigonometric Equations and Inequations Exercise 8:, Questions Asked in Previous 10 Years' Exam, (i) JEE Advanced & IIT-JEE, jr 1, , 140. Let S = x e (- it, it): x £ 0, ± — k The sum of all distinct, , 2J, , solutions of the equation -Ji sec x + cosec x, , + 2(tan x - cot x) = 0 in the set S is equal to, [Single Correct Option 2016 Adv.], i \, , 771, , (a) —, (c)0, , 142. For x E (0, it), the equation sin x + 2sin 2x - sin 3x =3 has, , (a) infinitely many solutions, [Single Correct Option 2014 Adv.], (b) three solutions, (c) one solution, (d) no solution, , 143. Let 0, <j>e [0,2rc]be such that 2 cos 0(1- sin 0) = sin2 0, , /, , g, , g\, , tan - + cot - cos <f> -1, tan (2tc - 0) > 0, I, 2, 2), , -J3, , 141. The number of distinct solutions of the equation, , — cos2 2x + cos4 x + sin4 x + cos6 x + sin6 x = 2 in the, 4, [Integer Answer Type 2015 Adv.], interval [0,2it] is, , and - 1 < sin 0 < - —. Then, 0 cannot satisfy, 2 [More than One Correct Option 2012], /(h), L\ It <0<, ., 47t, , (a) 0 < 0 < —, 2, , 2, 3, ... 3n . „, (d) — < $ < 2n, , www.jeebooks.in, / \, , 471, , A, , 3K, , (C) y < 0 < —

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, , 71, , 144. If P={0:sin0-cos0 = V2cos0}and, , Q = {0:sinO + cos6 = V2sin0}be two sets. Then,, [Single Correct Option 2011], (b)Q<ZP, (d)P = Q, , (a) P c Qand Q-P * 0, (c)P(ZQ, , 145. The positive integer value of n. > 3 satisfying the, equation, , 1, , • I—, 71, sm, , In, , ■, L_, +—, is, 2n, 371, sin, sin, n, nJ, [Integer Answer Type 2011], , 1, , (, , It, , 7t i, , (a) HMT’2n], 6, , _ v r 7t, it 1 [ 5n, (b) 0,— O —,271, , 6J, 6., , nn, that 0 — for n = 0, ±1, ±2and tan0 = cot 56 as well as, 5, sin 20 = cos 40 is, [Integer Answer Type 2010], , 147. The number of solutions of the pair of equations, 2sin2 0-cos20 =0 and2cos2 0-3sin0 =0 in the, [Single Correct Option 2007], , L6, , , . F 1t, 7t 1 f 2lt n, (c) 0, — u — , 2it, 3J L 3, 3., (d) None of the above, , (ii) JEE Main & AIEEE, 149. If 0 < x < 2n, then the number of real values of x, which, , satisfy the equation, , 146. The number of values of 0 in the interval —, — such, I 2 2J, , interval [0, 2n] is, (a) 0, (b) l, (c) 2, (d) 4, , 175, , cos x + cos 2x + cos 3x + cos 4x = 0, is, [2016 JEE Main], , (a) 3, (c) 7, , (b) 5, (d) 9, , 150. The possible values of 0 G (0, it) such that, [2011 AIEEE], sin (0) + sin (40) + sin (70) - 0 are, it 5n it 2n 3n 8n, . . 2n n 4ti it 3n 8n, (b)-,—, (a), 9’4’ 9’2’ 4 ’ 9, 4 12 2 3, 4 9, , . 271 it It 271 371 3571 ...... 27t 7C 71 271 371 87t, .4 ’ 9, , (c) —,, ,----- (d) —, -, —, 94234 36, 9423, 9, , 151. The number of values of x in the interval [0, 3n], satisfying the equation 2 sin2 x + 5sinx-3 = 0,is, , 148. The set of values of 0 satisfying the inequation, 2 sin2 0 - 5 sin 0 + 2 > 0, where 0 < 0 < 2n, is, , (a) 6, (c) 2, , (b) 1, (d) 4, , [2006 AIEEE], , [Single Correct Option 2006], , www.jeebooks.in

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www.jeebooks.in, Answers, , Exercise for Session 1, t__ ,A_. IX n, ,A_ on, l.x=(4n+l)—,-(4n-l)—, 14 ', '6, , 2. 3, , 3. 15, , 4. M7t± —, 8, , 6. (6n+l)—, 7.4, 8. nn± —, 12, 3, .. n, n, 10. (2n+ 1)—, nn± — 11. No solution, ' 8, 3, ,, n, 5n, 14.4, 13.— or —, 15.0, 12, 12, , 5. nn± —, 8, 9.—, 3, , 12. mt+ —, 4, , Exercise for Session 2, 1. 2nn or (4n+1)—, 2, , . nn n, -nn , n, 2. — + — or----- + —, 3, 18, 2, 6, , 4. 2nn- —, 4, , 5. 2nn+ — + —, 4 12, , 7. nn+tan"'2ornn+tan, , 3.2nn+ —, 3, , 61. (c,b) 62. (a,b,c), 66. (a,c,d), 71. (a), 72. (a), 77. (a,c) 78. (b,c), 83. (a,b,c,d), 86. (b), 87. (d), 93. (d), 92. (b), 98. (b), 99. (a), 104.(1) 105. (3), 110. (2) 111.(0), 116. (c) 117. (c), 121. (i) —>(2), (ii) No, , 63. (a,d), 64. (c,d) 65. (b,c,d), 67. (a,c), 68. (a,c) 69. (c,d) 70. (c), 73. (a,d), 74. (a,b) 75. (a,c) 76. (a,b), 80. (b,c) 81.(a,b) 82. (a,b,c,d), 79. (a,c), 85. (a,b,c,d), 84. (a,b), 90. (a), 88. (c), 89. (a), 91. (c), 94. (a), 96. (c), 97. (a), 95. (c), 102. (1) 103. (8), 100. (8), 101. (2), 108. (2) 109. (5), 107. (3), 106. (4), 114. (d) 115. (b), 112. (d), 113.(b), 119. (a), 120. (b), 118. (a), solution, (iii) 4, (iv) 0, (v) 3, , 6. 8, , ,,,, ~_r-3 r, 122. a e —andx= — +, , 8. nn, nn—, 4, , 123. a = 2nn ± —, n e Z, 3, , 3. (2n+l)—, , 124. x = nn + (-1)" — and y = 2mn ±, m,n e Z, 6, 3, 7t, 125. x = (2m + 1)— and y = 2nn + — ,nel, , 5. 0, , 126., , L2, |_2 2J, , 2, , It, , jr, , Exercise for Session 3, ,, 7n, 1. 2nn + —, 4, , 2. 2nn+ —, 6, , 4. A = nn± —; B = nn± —, 12, 6, 8. 3, 9. 5, 7. nn, , 6. No solution, , 10.0, , n . . 7t, .., 1. -n,- — u —, 6, 6, , '5n, _ fn) L nl f5n, l2j L 6j L. 66 ., , 3. — ,2n u{0}, , 4., , 5., , fo,-), I 4J, , J, , 6., , 7. 7?-|x:x=^^± — ,neZ, I, , 2, , r_, n n "i, L 6’iJ, , (, , [, I, , io’^oJ, , 19. (b), , 25. (b), 31. (d), 37.(a), 43. (b), 49. (a), 55. (c), , 2.(d), , 8. (b), 14. (a), 20. (c), 26. (b), 32. (a), 38. (a), 44. (a), 50. (b), 56. (b), , 4., 10., 16., 22., , (b), (c), (c), (a), , 39. (b), 45. (a), 51. (a), , 28., 34., 40., 46., 52., , (d), (c), (b), (b), (a), , 57. (a), , 58. (d), , 3- (a), 9. (c), 15. (d), , 21. (b), 27. (c), 33. (a), , J, , 6j, , j ancj, , con£jjtiOn is 0 < a < 1, , 133. a = 3n(4n + 1), , H. (a)., , 6. (c), 12. (a), , 17- (b), 23.(a), 29. (a), 35. (a), , 24. (b), 30. (b), 36. (c), , 41. (a), 47. (b), 53. (a), , 42. (b), 48. (c), 54. (a), , 59. (b), , 60. (c), , 5. (b), , 128. x = kn + —, y = mn and z = nn where k,m,n eZ, 4, 2ti, 2n, 129. x = 2mn ± — and y = 2(m - n)n ± —, m, n e I, , 131. x= 2nn± — + —, 4, 4, 132. x = nn + (-1)" sin'1, , Chapter Exercises, 1(d), 7. (c), 13. (d), , 3, , 130. 0 = 4nn± —, 3, , n 37t'\, , 8.1 n7t,n7t + — I, , 4, , 2n, \, 2,2nn± — -2, , 71, 127. x = -1 and y= nn± — +1, n e /, 4, , Exercise for Session 4, , .2, , (, , I, , sin~‘(l- >/2a+ 3), , 2, , 18. (b), , 134. x = — ±(-l)"2, 8, 7C, 135. x = y=mn±—,nel, 4____, 136. x = nn ± cos'1 Va+ 3, where n e z and a e[- 3, -2], ft, 137. x = 3, , ., , 71, 71, n, n, 139. x = 2mn + —, 2mn, nn ± — and y = kn\ m, n,k,el, 2, 6, 144. (d), 142. (d), 140. (c) 141.(8), 143. (a.c.d), 147. (c), 149. (c), 145. (7) 146. (3), 148. (a), 150. (a) 151. (d), , www.jeebooks.in

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www.jeebooks.in, 178, , Textbook of Trigonometry, , -7t It, 17. 0e —,- =>-i<sin0<l, 2’2., , 11. Consider R.H.S i.e., 5’+5"r, 5‘ + 5~*, , >(5X-5’I),/2, , But here 0<sin0<l, , 2, 5r+5'r>2, , =>, , [By using A.M > G.M], , => From the given equation sin(e* )>2, , which is not possible for any real values of ‘x’. Thus, the given, equation has no solution., , Now, logun0cos2 0=2, [By using loga b=c => b-a'], cos2 0=sin2 0 =J tan2 0=1, -7t n", 0=n7t + —V0e, 4, 2 2, , 12. We know that in a triangle A + B + C=180°, , 0=-, 0=—, 4, 4, , and 2B=A + C (As A, B. C are in A.P), , =>, , [As logaxis define for a>0 or 0<a<l], , 3B = 180 or B=60°, , (Reject as sin, , Now ‘0’ be the common difference between A, B and C,, , C-A=20, , then, , ...(i), , sin(C-A)=l/2, , (given), , sin(20)=l/2, 571 „ 7t, 57t, 71, 20=— or— => 0=—or —, ", 12, 6, 6, 12, „ „, 7t n, 7t 5tt, A=B-B-------- or, 3 12, 3 12, , Now,, , . _, , 71, , [Using Eq. (i)], , 71, , cos 0, =n, /■I, , =>, , (2sinx-l)>0, /rc 5ti, . . 1, —J,sinx>—- — =>xe —,—, 2, 6 6, , ^sin0(=O, , Case II: If n>2, sin" x and cos" x both decrease then, sin" x+cos" x<l, (as 0<x<7t/2), , [As sinx+2>0VxeR ], , =>, , Case III: If n<2, sin" x and cos" x both increases then, sin" x+cos" x>l, (asO<x<7l/2), Then, sin" x+ cos" x>l for n<2, , =>n €(-«»,2], , Consider x2 -x-2<0, (x-2)(x+l)<0 => -l<x<2, , -(ii), , From Eqs. (i) and (ii),, 571, , ,, Now. as, 2<—, we obtain that ‘x’ must lie in -.26, 6 J, , k, , 25, , 5, , 5, , \2, , 14, , =>(i) becomes sinx+cosx=l, • ( ., 1_, sin x+ — =, , 15. Wehavea1+a2cos2x + a,sin2x=0, , =>, , x=n7t + (-l)" —, 4 4, 21. sin7 y =|x3 - x2 -9x+9| + |x3 - 4x- x2 + 4|+sec2 2y + cos4 y, (According to the choices), sin7 y =sec22 2y + cos4 y, , => which is zero for all value of ‘x’., k, , Ci, , sin’y -cos22y=1 + cos4y'cos22y, , l., , If a, =-^-=-a2 or a, =—, a2 =-, a, =k, Lt, , ii, , Now, for x=l, , =>a, + a, cos2x+a3(l -cos2x)/2=0, ~k, , 47, , x+ —=n7t + (-l)" —, 4, 4, , o, , + —>2, ', 5, , As y =2sinx<2, so there cannot be any point of intersection., , flj, , •(i), , aeR, , I, , f, 1, 1, 3 1, x+- \j +------= 5 x+5, , 20. sinx+cosx=min(l,a2-4a + 6}, As, a2 -4a + 6=(a -2)2 + 2>2 for all a, , 2, 3, 14. Consider :y=5x2+2x+3=5 x2 + -x+5., 5, , ir, , i, , 19. Case I: For n =2,sin2x+cos2x=l., , (2sinx-l)(sinx+2)>0, , =5, , n, , 18. Consider, , n, , 571, , 71, , A = 45°, B=60°,C=75°, , =>, , => The given equation has unique solution., , COS0, =1, cos02=l, COS0j =1,..., cos0n =1, 0^ = 0,=..-0n=O, , 13. Consider : 1st equation i.e. 2sin2 x+3sinx-2>0, =>, , V2, , cos 0, + cos 02 + cos 03 +...+ cos 0„ = 1 +1 +1...+ n is valid only, when, * n~times ', , =>A=— as a cannot be less than ‘0’ and C=— + — =, 4, 3 12 12, =>, , -A<o), , Lt, , Now, R.H.S >1 and LH.S <1, , For any keR, , =>, , Hence, the required number of triplets is infinite., , =>, , sin7y-cos22y=l, , =>, , sin’y=l and cos22y=l, n, , 16. We have, sin4 x-(k+2)sin2 x-(k+3)=0, sin2 y=(k + 2)±A/(k + 2)2 + 4(k+3) = (k + 2)±(k+4), SU1 X, 2, 2, , Either sin2 x=k+3 or sin2 x=-l, , L.H.S =1, , _, 7t, General values of ‘y’ is 2nn + —, 2, ,, „, 7t, Hence,, x=l andy=2n7t + y, , www.jeebooks.in, 0<sin2 x<l or not possible, , 0<k+3<l, , -3<k<-2

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www.jeebooks.in, 186, , Textbook of Trigonometry, , n 3rc, 85. Atx=—, 2 2, , Sol. (Q. Nos. 91 to 93), 91. v ZOMN = 15° = ZONM, , [l+sinx]=0, [l-cosx]=l, sinx=0 + l =>-1=1, , (Absurd), , (1+sin x)=l,(l-cosx)=0, sinx=l + 0=>0=l, , (Absurd), , [1 + sin x]=2-[l-cosx]=l, sinx=2+l=3, , (Absurd), , sinx=l + 2=3, , (Absurd), , .-., , ZAfON = 180°-15°-15° = 150°, , At x=0, , 71, , At x=—, 2, , At x=7t, n, In —,0 , [l + sinx]=0, [l-cosx]=0, 2, sinx=0 + 0=0, (, , 92. Z.PLO = Z.PMO = 90°, (Absurd), , 71 j, , In 0,— ,[l+sinx]=l,(l-cosx]=0, k 2J, sinx=l + 0=l, , Now, quadrilateral ONRM is cyclic, [•/ ZOMR = ZONR = 90° ], ZR = 180° - 150° = 30° = ZQ, ["/ PR = PQ], =>, ZP = 120°, , (Absurd), , /. Quadrilateral PLOM is cyclic and OP is diameter of, circumcircle, =>, Z.LOM = 180° -ZP = 60°, =>, Z.POM= 30°, , In I —,7t , [l + sinx]=l, [l-cosx]=l, , 12, , J, , sinx=l + l=2, , (Absurd), , In n,— L [l+sinx]=0, [l-cosx]=l, , I, , 2/, , .’., sinx=0+l=l, /. All the four results hold., , (Absurd), , Sol. (Q. Nos. 86 to 88), 3sin2x-7sinx + 2 = 0, sinx = - or 2(Reject), 3, , 86. N = 2, , Now, is right angled A POM,, [•/ AC = x sec 9], , OP = — sec 30°, 2, _73 2, -r = l, 2 <3, , One value in first quadrant and other lies in second quadrant., , P, , 87. Let x + ct, then two values a and n - a., =>, sum is n, oo, -l, 2^2, 88. •/ sin a = - => cos a = ±----3, 3, T4(a) = sin4 x + cos4 a = 1 - 2 sin2 a cos2 a = ~, , Sol. (Q. Nos. 89 to 90), 89. If a = 3 and a =5, then N G [55, 54], , Radius of circumcircle = 2, , largest value of N is 54 - 1 = 624, smallest value of N is 55 = 125, Difference of largest and smallest integral value of N, = 624-125 = 499, 90. If a = 2 and a = 2, then, N, is(23) -(22) = 4, If a = 1 and a = 3 then N2 is 32 - 31 =6, y = (N, sec26 + N2 cosec2 0), = (4 sec20 + 6 cosec20), /.The minimum values ofy is(2 + V6)2 = 10 + 4^6, , 0, , <3, 2, , M, , Area = rc, , 1, 2, , 2, , n, 4, , 93. •/ ZOQN = 15°, ON = —, 2, 0, , www.jeebooks.in, V3, 2

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www.jeebooks.in, Chap 02 Trigonometric Equations and Inequations, 1, , . „, n 2n + l, sin 0+cos 0=-----2, . ,, 2n + l, cos(n/4-6)=^-, , Since, , 115. In statement,, , "1<---- 7-<l, , 2V2, =>n=0 or -1 as n is an integer, cos(n/4-0)=± (1/2-72), =>, =>, , 8cos2(n/4-0)=l, , =>, , 16cos2| — -01=2, , 6tan(x/2)+4-4tan2(x/2) = 5+5tan2(x/2), 9tan2(x/2)-6tan(x/2)+l = 0, 90tan2(x/2)-60tan(x/2) = -10, , =>, =>, , 90 tan2--60 tan-+ 10 = -10+10 = 0, 2 2, , 112. When n=l,. we have interval [0,7t ], which covers only the first, and second quadrants in which., sinx=-l /2 is not possible. Hence, the number of solutions, zero. Also, from2(n-l), we have zero solution when n=l., For n=2, we have interval [0,2n] which covers all the quadrants, only once. Hence, the number of solutions is two., Also, from2(n-l), we have two solutions, when n=2., Fro n=3, we have interval [0,37t], which covers the third and, fourth quadrants only once. Hence, the number of solutions is, two. But from 2(n-l), we have four solutions which contradict, Hence, Statement I is false, and Statement II is true., , => (sinx+2cosx)(cosx-sinx)2=0, => (sinx+2cosx)(cosx-sinx) ’=0, which does not hold for any value of x as -n 14<x<n 14, _ 1, 116. For (n/4,71 / 2), O<cos0<-7=<sin0<l, V2, (cos0)"*9<(sin0)"*9, So,, (cos0)’u,9<(sin0r9, * and, (cos0)"9+(cos0r9 <(sin0)™9, Showing that Statement I is true., In Statement II let e'“e=r, t2-4t-i=o, Then,, =>, , =>, , => sin0>l which is not possible., So, the give equation has no solution and the statement II is, false., , 117. sin2x+sin4x+sin‘x+...inf, sin2 x, ,, = -———=tan x, 1-sin x, , => Statement II is false., , y=lxi, , \ /, , Now, exp {tan2 x log, 2} =2*“ ’, , y=isinxi, , X'-, , 2, e“>9=2±45 =»in0=log(2± V5), , Since, 2—75 <0, sin0=log(2+ -^)>log,, , 113. The graphs ofy=|sinx| and y=|xj is, y, , 1, -1, , sinx cosx = 0, 1 cosx sinx, , 111. From the given equation we have, 3 2tan(x/2), l-tan2(x/2), l + tan2(x/2) l + tanz(x/2), =>, , cos a, cosa sin a = 0, sin a -cosa, , 1, , J, , 14, , sin a, , => either sin a=0 or tana=l, =>, a=n/4(as0<a<n), and Statement n, (sinx+2cosx), 1 cosx cosa, , -l<cos(n/4-0)<l, 2n + l, , =>, , 189, , So2t“‘x =1 or2t“’x =8, X, , 2‘“’x=8, , (v tanx>0=^2“’x>l], , tan2 x=3 => tanx=V3, Y, | sin x |=| x | has only one solution x=0. But Statement II is not, the only explanation of Statement I., j., . „, 2tanx, l-tan2x, 114. sm2x=------- —, cos2x=------- —, 1 + tan x, l+tan2x, , => Statement II is correct., In Statement I, we have, , cos2 x= sin2x, cosx(cosx-2sinx)=0, tanx=l/2 as cosx*0, From Statement II we get, 1+1—l/4_7, sin2x+ cos2x=, 1+1/4 ~5, , Now,, , cosx _, 1, cosx+sinx 1 + tanx, , 1 Ji-i, ~Ji+i 2, , -V3, 1, 118. (A)sinx=—and cosx=----2, 2, In o lln, In, 5n, a=—, p=— a=—, v=—, 6, , 6, , 6, , 6, , lln 5n, , o, , p-y--------- =n, 6, , =>, , 6, , o In lln „, a+B=—+—=3n, 6, , 6, , www.jeebooks.in, Statement I is false, , A q,s, (B) cot x= —Ji and cosecx=-2, , lln o 5n, , 7n, , a=—, P=—»Y=—, 6, , 6, , 6

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www.jeebooks.in, 190, , Textbook of Trigonometry, a-{3=n andP+Y=2n, , 2cosx + 1 „, 1, = 0 => cosx = —, 2, sinx, , =>, , ^p.t, , (C) sin x=--and tanx=-7=, 2, y/3, 771 o 1171, It, a=—,p=--- ,y=6, 6, 6, a-Y=7t, a+P=37t, P+y=2ti, a C—>r,s,t, 119. (A)2sincos0=l/V2 if cos0>O, =❖, , sin20 = l/V2 => 0 = 7t/8 or371/8, sin20 = -I/V2 if cos0<O, , =>, 0 = 571/8 or 771/8, A A->p,q, (B) 2cos20cos40+ cos40 = 0, (2cos20+l)cos40 = 0, Either cos 40=0, n 71 371 571 771, =>, 0 = —, —,--- ,--- or cos20 = -l/2, 8 8 8 8, =>, 0 = 71/3 or 27t/3, /.B—»p,q,r, (C)(4cos2 0-l)(2sin0-l)=O, cos2 0=1/4 => cos0=±l/2, =>, 0=71 Z3,2n/3 or sin0=1/2, 0=71/6, .’.C->r,s, , 120., , 7t 371 57t 7tt, (D) If sin 40=±1, 0=—, 8’ 8 ’ 8 ’ 8, .'.D—*p,q, sin0 + sin30 + sin50+ ...+ sin((2n -1)0), £(0) =, cos0+ cos30+ cos50+ ••• + cos[(2n-l)0], (n —1)20, sin-(20), 0+, 2, 2, sin, . 20, sm —, sin(n 0), ______ 2, = tann0, (n -1)20", cos(n0), 0+, sin ~(20), 2, cos, . 28, sm—, 2, , => x = 2mt ± —, n e I and 0 < x < 2n., 3, 271 471, X = —, ■—, , 3, , 3, , (ii) We have, sin’ xcosx + sin2 x.cos2 x + sinx.cos’ x = 1, =>, sin x cos x(sin2 x + sin xcosx + cos2 x) = 1, sin2x, sin2x, 1 +------ = 1, =>, 2, 2, sin2x(2 + sin2x) = 4, sin2 2x + 2sin2x - 4 = 0, -2±J4 + 16, r, =>, sin2x =---------------- = -1 ± V5, 2, This is not possible, as -1 sin 2x < 1., Hence, the given equation has no solution., (iii) We have, 2C0" = |sinx|, It is true only for cosx = 0 and | sinx | = 1, =>, , => cosx = cos—andsinx = ± 1 =sin| ± —, , V 2), , 2, , =>, , x = 2nit ± —, 2, But, x e [-271, 2ti], n n 3tc 371, x - —, —, — ,----- Hence, number of solutions = 4., 2 2 2, 2, (iv) We have,, | cosx| =[x] =y (say), , y, , y=[x], y=lsinx|, , *x, , X'-, , r, Graph of | cosx | and [x] don’t cut each other for any real, value of x., Hence, number of solutions is zero., (v) We have x + 2tanx = —, 2, it x, or, tanx = —, , (a)4^=, (B)/<, , Manx, , 2 -1, , (C)A, X'(D)/,, , 2-V3, , 121. (i) If cotx > 0, then —^— = 0 which is not possible,, sinx, Now if cotx < 0,, 1, then,, -cot X = cot x +----sinx, , 2, i, i, i, i, i, i, i, i, , I, I, I, I, I, I, , I, I, I, I, I, I, I, I, I, I, I, I, , 3k, 2, , i, i, i, i, i, i, i, i, / i, ' i, i, i, , 2n, , 5k, , ?, , i, , <1 I, I, I, I, I, I, , •X, , I, I, I, I, I, I, , www.jeebooks.in, r, , /, n X, y=4~2

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www.jeebooks.in, 192, , Textbook of Trigonometry, , = 13 + 4cos2(x + y), => 4cos2(x + y) + (6-74x - x2 -8)cos(x + y), +(13 -6>/4x- x2) = 0 ...(i), , Let, cos(x + y) = t and 6^4x - x2 - 8 = a => 6^4x - x2 = 8 + a, Clearly,, 4x - x2 £ 0 or x(x - 4) £ 0 or 0 < x, Now,, equation (i) reduces to;, 4t2 + at + (5 - a) = 0, which is quadratic in t., D£0, a2 - 4.4(5 - a) £ 0, a2 + 16a -80 2: 0, (a + 20)(a - 4) > 0, =>, a < - 20 or a > 4, However, according to substitution;, a = 6^4x - x2 - 8, , 4, , ...(ii), , ...(iii), , -8 < a < 4, , •••(iv), {from (iii) and (iv)}, , a=4, Hence, 6^4x-x2 -8 = 4, , ^4x - x2 = 2, 4x - x2 = 4, x2 - 4x + 4 = 0, (x - 2)2 = 0 or x = 2, equation (i) becomes;, 4cos2(2 + y) + 4cos(2 + y) + 1 = 0, (2cos(2 + y) + I)2 = 0, 2cos(2 + y) + 1 = 0, cos(2 + y) = -1 = co: 7t----ft'), , 3J, , „, , „, , . 2lt, , 2 + y = 2mt ± —, y = 2mt ± — - 2, n e I, 3, , (, , 2n, A, Thus, the solutions are 2,2mt ±------ 2, , I, , cos2 y > 1, which is only possible when,, cos2 y = 1, Putting cos2 y = 1 in second equation we get,, sin2z = 0 =>z = mt, ...(ii), Similarly, cos2 y = 1 => y = rmt, ...(iii), Alos,, tan2 x + cot2 x = 2 cos2 y, =>, tan2 x+cot2 x = 2 => tan2 x = cos2 x = 1, , 3, , J, , x = kit ± —, y = rmt, z = mt where k, m, n 6 Z, 4, , 129. The given equation can be rewrite as, x+ y, x+ y', 2 co:, 2, 2 >, x+y, x~y + 1 = 0, or 4 cos2, + 4 cos, 2, < 2, \\2, x-y, or, + cos, + sin2 x-y = 0, 2, 2, x + y') _, J-1 and sin2 x-y = 0, => 2 co:, 2 )~ -cos 2 ), 2, , H+), , from second equation, we get,, x - y = 2nit, where n 6 I, or, y = x - 2mt, Substituting the values of y in the first equation, we get, 2cos(x - nit) = -cos mt., =>, 2cosx. cos mt = -cosmt, 1, n, , 2it, it, => cosx = — = co: it — => x = 2zmt ± —, me I, 2, 3, 3, Solutions are,, , 127. Rewriting the given equation as,, {tan(x + y) - cot(x + y)}2 = - (1 + 2x + x2) = - (1 + x)2, or {tan (x + y) - cot (x + y)}2 + (1 + x)2 = 0, which is possible only when,, tan (x + y) - cot (x + y) = 0 and 1 + x = 0, , tan2(x + y) = 1 = tan2 — and x = -1, , „, 2n W, , 2it, 2 rmt ± — , 2(m - n)it ± —, 3, 3 J k., 0, 2, , 0, 2, , 130. Here cot---- cot 0 = cosec —, 9, , =>, , cos—, COS0, ___ 2 --------= cosec, , . 0, 2, , Now, tan2(x + y) = tan2 — => x + y = mt ± —, ,, , y = mt ± — + 1, n e I, 4, i, , 7t, , The required solutions are 1-1, mt ± — + 1 n e I., , =$, , 0, 2., , sinO, , sin —, , , it, , ..J.(iv), , x = kit ± —, 4, , Hence, the solutions are, , = 6-^4—(x — 2)2 -8, , Now,, , 128. As we know, A.M. £ G.M., tan2x + cot2x £ -Jtan2 x. cot.2 x, 2, =>, tan2 x + cot2 x > 2, Now, from the first equation,, 2cos2y >2, , 0, , 2 cos2---- cos0, 0, 2, = cosec —, sin0, , =>, , 2, , 0, 2cos2- -| cos2 - - sin-0, 2 - I =sin0. cosec -0, 2, 2, 2, 2J, , www.jeebooks.in, =$, , ., , 9, , • n, , 1 = cosec —. sm0, 2

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www.jeebooks.in, 196, , Textbook of Trigonometry, , 5x, X, COS X ■ cos----- cos — = 0, 2, 2, 5x, X, cos x = 0 or cos — = 0 or cos — = 03, 2, 2, cos x = 0, 71 371, X = —, —, 2 2, , 146. Given, tan 0 = cot 50, , ( It, tan0 = tan-----50, k2, , Also,, , => --50 = 071 + 0, 2, , 60 = —-nn => 0 = —, 2, 12, (it, cos4 0 = sin20 = cos-----20, , nil, , Now,, , 6, , U, , 5x, cos — = 0, 2, 5x 7t 37t 5n 7n 9n 117t, 2’ 2 ’ 2 ’ 2 2’2, 2, 7C, .. 371, ..., 7 it 9n, [v 0 < x <2n], x = —, —, it, —, —, 5’ 5, 5 5, 71 371 571, cos - = 0 => - =, 2’2’2, 2, 2, , I 71, 40 = 2nn±--20, , 12, , =>, , Taking positive sign,, nil, 60 = 2nn + — => 0 = —+ —, 2, 3, 12, , Taking negative sign,, , and, , it, n, it, 20 = 2rt7t---- => 0 = ZlTt------2, 4, , Above values of ©suggest that there are only 3 common, solutions., , Hence,, , sin 4 0 + (sin 0 + sin 7 0) = 0, sin 40 + 2 sin 40- cos 30 = 0, , 2cos20 = 3sin0 => sin0 = 2, , =>, , => Two solutions exist in the interval [0, 2tc]., , 40 = it, 271,371, 1, cos 3 0 = 2, „ „n, 27t 471 87t, O<30<37t => 30 = —,—,—, 3, 3 3, q, 71 71 371 271 471 87t, ”4’2’ 4 ’ 9 ’ 9*9, 9, , [where, (sin 0 - 2) < 0, V 0 e /?], (2sin0-l)<O, , Y, , y-i, O £, 6, , 5n, 6, , K, , 2k, , X, , 151. Given equation is 2sinz x + 5sin x - 3 = 0., =>, , Y', , sin 0 < 2, :., , it, From the graph, 0 e 0, —, 6, , 0 < 0 < it, O<40<4n, , (2 sin 0 - 1) (sin 0 - 2) > 0, , X', , sin 4 0 {1 + 2 cos 3 0} = 0 => sin 40 = 0, cos 3 0 = -, , As,, , 148. Since, 2sin2 0-5sin0+2>O, , =>, , [v 0<x<2n], x = it, 7t 3. tc 7tt 971, It 371, X = —, —, 71, —,--2 2, 55’ 5 ’ 5 ’ 5, , 150. sin 0 +sin 40+sin 70 = 0, , 147. 2sin20-cos20 = 0 => sin20 = 4, Also,, , 0<x<2n], , (2sin x - l)(sin x + 3) = 0, 1, sin x = 2, Y, , 571 „, —,271, 6, , [v sin x # - 3], , y = sin x, , 149. Given equation is, , cos x+ cos 2x + cos 3x + cos 4x = 0, , O n/6, , 2k, , 3n, , y= 1/2, X, , (cos x + cos 3x) + (cos 2x + cos 4x) = 0, , 2 cos 2x cos x + 2 cos 3x cos x = 0, , 2 cos x (cos 2x + cos 3x) = 0, I, 5x, x 1, 2 cos x 2 cos — cos — = 0, , V, , 2, , It is clear from figure that the curve intersect the line at four, points in the given interval., , Hence, number of solutions are 4., , 2J, , www.jeebooks.in

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www.jeebooks.in, CHAPTER, , Properties and, Solutions of, Triangles, Learning Part, Session 1, • Basic Relation between the Sides and Angles of Triangle, Session 2, • Auxiliary Formulae, Session 3, • Circles Connected with Triangle, Session 4, • Orthocentre and its Distance from the Angular Points of a Triangle and Pedal Triangle, • Centroid ofTriangle, Session 5, • Regular Polygons and Radii of the Inscribed and Circumscribing Circle a Regular Polygon, Session 6, • Quadrilaterals and Cyclic Quadrilaterals, Session 7, • Solution of Triangles, Session 8, • Height and Distance, , Practice Part, • JEE Type Examples, , • Chapter Exercises, , www.jeebooks.in, Arihant on Your Mobile!, Exercises with the S symbol can be practised on your mobile. See inside cover page to activate for free.

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www.jeebooks.in, Session 1, « ■ >«■!, , ■ ■ In* «-!■ I, , .«■—w<, , T^,r.a», , T 'J-.-, , :’P '■! »!■—.— u, , WIWWIM >■»»■» niipx 1W ——■«», , Basic Relation between the Sides and Angles, of Triangle, Basic Relation between the, Sides and Angles of Triangle, , A, , From AACD,, , In a AABC, the angles are denoted by capital letters A, B, and C and the lengths of the sides opposite to these angles, are denoted by small letters a, b and c, respectively., Semi-perimeter of the triangle is written as :, a +b+c, s =---------- ,, 2, , sin(180°-C), sm(180°-C) =, =—, ---- = —, AC, b, AD = £>sin(180° - C) = bsinC ...(iv), , c., , b, , From Eqs. (iii) and (iv), we get, ...(2), , csinB = bsinC, , JSO’-c, , Some geometrical properties of A,B,C and a,b,c. In AABC, (i) A + B + C = 180°, , A, , b, , [•/ D and C are same point], (iii) a>0, b>O,c>O, , Sine Formula or Sine Rule, , a, , &sinC = csinB, , [•.'C=90°] ...(3), , Similarly by drawing perpendicular from from C to AB, we, can prove that, a, b, -(5), sin A sinC, , Proof: Case I. When ZC is acute :, A, , From A draw AD ± BC, , C(D), , Thus from (1), (2) and (3), it follows that in all cases, b, c, bsinC = csinB or, sin B sin C, , In any AABC, the sides are proportional to sines of the, opposite angles,, a, b, c, i.e., sin A sin B sin C, , 't', , In AABD,, B, , AD = csinB, ., AD AD, In AACD, sin C =----AC~ b, AD = bsinC, , From Eqs. (i) and (ii), we get csin B = bsin C, , From (4) and (5), we get, a, , b, , c, , or, , or, , B, , or b = c sin B, or, , D, , a, , Case III. When ZC = 90°, Draw AD _L BC, AD AD, In AABC, sinB = —, c, AB, AD = csinB, or AC = csinB, , and its area is denoted by A., , B AD AD, sm B =---- =, AB, c, , C, , B, , D, a, , sin A, , C, >, , b, , c, , sinB, , sinC, , Cosine Formula or Cosine Rule, -(ii), , ...(1), , (i) cos A =, , !>2+ci-ai, , ora2 =b2 + c2 -2bacosA, , 2bc, , (ii) cos B =, , Case II. When ZC is obtuse :, , c +a — b, , 2ac, orb2 = c2 + a2 -2accosB, , From A draw AD ± BC, InAABD,, , a2+b2-c2, , www.jeebooks.in, o AD AD, smB =---- =, AB, c, AD = csinB, , (iii) cos C =, , 2ab, , orc2 =a2 +b2 -2abcosC, , ...(iii)

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www.jeebooks.in, 200, , Textbook of Trigonometry, , fcsin A - ksinB', , n BD, -cosB = —, a, , C, •cot —, ksinA + ksinB ;, 2, , C, , A+B, , A-B}, , C, cos —, ___ 2_, A + B', A-B}, C, sin —, 2 sin, • cos, 2, 2, 2 7, , 2 cos, , • sin, , 2 J, , 2, , b, , A, B^, A, , £, 90>, D, , B, , c, , sin, , ■, , A-B, , BD --acosB, , cos, , In ACAD, AD, cos A -----AC, a, , A-B, , and sin, , 2, A+B, , AD, , b, , = tan, , (C>, , ufc, , = sin —, , = cos, , 2, , 2, , 2, A-B, , = LHS, , 2, , AD = bcosA, , or, , A+B, , as, cos, , 2, or, , [using Eq(i)J, , Now, c = AB = AD - BD = bcos A + a cos B, , ...(ii), , tan, , A-B, , a -b, , C, cot—, ^a+b 7, 2, , 2, , Case III. When ZB = 90°, In ACAB,, , Similarly it could be shown,, 'B-C, b-c, C-A, A, tan, cot —, tan, b+c, 2, 2, < 2, , x AB .c, cos A = — = AC ib, , B, cot —, c+a, 2, c-a, , c = bcosA, , I Example 1. Find the angles of the triangles whose sides, are 3 + Jl, 2J3 and Jb., , = bcos A + ccosB, C, , Sol. Leta = 3 + Ji,b = 2ji,c = '6, , a, A, , =>, , 0 +c -a, , 12Ji, 1- ., , 2 Ji, , 12J2, , [vccosB = ccos 90'i° = 0, as cos 90° = 0], , 124-6 — (9 4-3 +, , 2bc, , 6 - 65/3, , B(D), , c, , cos A =, , cos A = cos (60° + 45°), , Thus in all cases, c = b cos A + a cos B, , as cos (60° + 45°) =, , Similarly, we can prove that, , i-Ji, , b = ccos A + £2 cos C, , and, , A = 105°, , a = bcosC + ccos B, , a, , b, , sin A, , sinB, , Applying Sine formula, , Tangent Rule or Napier's Analogy, In any A ABC, tan, , A-B, 2, , Proof: In AABC, we know, a, b, c, sin A, , sin B, , sinB = -sin A =, a, , a-b, C, ------- cot —, a+b, 2, , = k (say), , 2J3, , sin C, , rsin (105°), , {sin60°.cos45° + cos60° - sin 45°}, , sinB = 4= = sin45° [v B * 180 - 45° as B + A < 180°], J2, , Now, RHS, C, • cot —, a + b;, 2, , 2ji, , 2ji, ____, Ji +1, Ji(Ji +1)[ 2J2, , [sine law], , a = ksinA,b = ksinB,c = ksinC, a-b', , 2J2, , =>, Here,, =>, , B = 45°, A = 105°, B = 45°, C = 180° - (A + B) = 180° - (150°) = 30°, ZA = 105°, ZB = 45° and ZC = 30°, , www.jeebooks.in

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , I Example 2. The sides of a triangle are 8 cm, 10 cm, and 12 cm. Prove that the greatest angle is double of, the smallest angle., , Sol. Let a = 8 cm, b = 10 cm and c = 12 cm. Hence, greatest, angle is C and the smallest angle is A, {as we know great­, est angle is opposite to greatest side and smallest angle is, opposite to smallest side.} Here, we have to prove C = 2A,, applying cosine law, we get, a2 + b2 -c2 64 + 100- 144 1, cosC =, (0, 2ab, ~, 2-8-10, 8, , and, , cos A =, , 3, , 2-10-12, , 4, , -(ii), , 9, cos2A = 2cos2 A -1 = 2------ 1, 16, , I, , [using Eq. (ii)], , 1, , „x, , cos2A = 8, From Eqs. (i) and (iii), we get, cos 2A = cos C, C=2A., , -(iii), , a+b ,, ------ = k, 11, 12, 13, =>, 2 (a + b + c) = 36k, b + c = Ilk, c + a = 12k, a + b = 13k, On solving Eqs. (i) and (ii), we get, a = 7k, b = 6k, c = 5k, b2 + c2 - a2 36k2 + 25k2 - 49k2, Hence, cos A =, 2bc, ~, 60k2, c+a, , 12, , 1-2., , 60, , 5, , a 2 +. c - bL2, , i.e., AC = n, AB = n + l, BC = n + 2, Smallest angle is B and largest one is A., A=2B, Here,, A + B + C = 180°, Also,, =>, 3B + C = 180° => C = 180°-3B, sin A sin B sinC, Using, sine rule,, n+2, n, n +1, n+2, , n, , sin2B, , sinB, , n+2, , nn, , n +1, , (i), , (ii), (ii), , (iii), , ...(i), ..(ii), , sin3B, , 2sinBcosB _ :sinB, , n, , n+2, , =>, , n, , n+1, , sinB, , sinB{3- 4sin2B}, , n, , n+1, , n +1, , =>, , = 3- 4(1- cos2 B), , 49k2 + 25k2 -36k2, 70k2, , n+1, , ...(v), , ( n + 2^, = -11 +: 41-----V 2n ), , l±A + i =, , 49k2 + 36k2 -25k2, , ■(iv), , n, 221 = -1 + 4cos2B, , 'n2 + 4n + 4', , 7, ", , n, , \, , 2n +1, n-, , n2 + 4n + 4, , 84k2, , 2a b, 60 5, 84 " 7, , n +1, , n +2, cos B =----2n, and from Eqs. (ii) and (iii);, sinB 3sinB - 4sin3B, , n, , a2 + b2 - c2, , B, , n+2, , n, From Eqs. (iv) and (v), we get, , 35, , 2ac, 38 19, 70 ~ 35, cosC =, , c, , From Eqs. (i) and (ii);, , "TF", , cosB =, , Sol. Let the sides be n, n + 1, n + 2, , sin2B_sinB_, sinB, sin(180°-3B), , I Example 3. With usual notations, if in a AABC,, b + c c + o a+b, then prove that, 11, 12 =7T, cos A COSB COSC, ~25~', , Sol. let,*±£, , I Example 4. The sides of a triangle are three, consecutive natural numbers and its largest angle is, twice the smallest one. Determine the sides of the, triangle., , b2 + c2- a,2, 2bc, 100+ 144- 64, , 201, , 7, , n, 2n2 + n = n2 + 4n + 4, , 25, , www.jeebooks.in, =>, , cos A, , 35, cosB, , cosC, , 7, , 19, , 25, , n2 - 3n - 4 = 0 => (n - 4)(n + 1) = 0, , n = 4 or -1, n = 4. Hence, the sides are 4, 5, 6., , where n * -1, , *

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www.jeebooks.in, 202, , Textbook of Trigonometry, , a+b+c, , I Example 5. Let 0 be a point inside a AABC, such, that ZOAB = ZOBC = ZOCA = w. Then, show that, cot w = cot A + cotB + cot C., , 2, ., 2C, ,B k, bcos —+ ccos — = —, 2, 2, 2, , Sol. In AOAC, Using sine rule,, sin(A - w) _ sin(180 - A), , A, 2 C 3b ,,, I Example 7. In a AABC, ccos2 — + acos — = —, then, 2, 2, 2, showa,b,c are in AP., , b, sin A, ~b, , OC, sin(A - w), OC, , e . in L., 2A, 2 C 3b, Sol. We have, ccos — + a cos — = —, 2, 2, 2, , Also, in AOBC, A, w, , c,, , O,, , => |(1 + cos A) + ^(1 + cosC) =, , X\, \ Czo, , => a + c + (ccosA + acosC) .= 3b, [using projection formula], =>, a + c + b = 3b, =>, a + c = 2b, which shows a, b, c are in AP., , ,b, , 180°w, , B, , C, , a, , sinw, , sin(180 - C), , ~6c, , a, , 7t, , I Example 8. In a AABC, o = 2b and |A - B| =y., Determine the ZC., , sinw, , sinC, a, OC, ..... sin(A - w), On dividing Eqs. (i) and (ii);, sinw, i, , (ii), , Sol. Given,, , a = 2b, ZA > ZB, , asinA, , BsinC, , a, , c, , as,we know,------ =------- =, =k, sin A sin B sinC, , sin(A - w) ksin A - sin A, sin w, ksinB-sinC, sin A cos w - cos A sin w, {sin(7t - (B + C))}, = sin A, sinw, sin Bsin C, sin B cosC + cos Bsin C'], => sin A(cotw) - cos A = sin A, sin Bsin C, J, , =>, , o, , or, , ...(ii) [as A > B], , . I—, 71 I = 2b - b cot —, tan, 16 J 2b + b U J, , [using Eqs. (i) and (ii)], , 1 1 , C_, —f= = - cot —, 2, y/3 3, , cotf—1 = 73, <2 J, C = *=*C=*, 2, 6, 3, , i.e., , the AABC., B, , 71, , A-B = —, 3, Using Napier’s analogy, we have, A - B^l a - b, I C, „, tan, -- =------ cot| —, 2 J a+b, 2J, , I Example 6. Solve, bcos2 y +ccos21 in terms of k, where k is perimeter of, , Sol. We have, Bcos2 — + c cos2 —, 2, 2, b., c, —(1 + cosC) + —(1 + cosB), 2', 2, b+c, + -(BcosC + ccosB), 2, 2, b+c 1, =>, + -a, [using projection formula], 2, 2, , •••(i), [as a > b and we know greater angle is, opposite to greater side], , |A-B|=^, , sin A(cotw) - cos A = sinA(cotC + cotB), cotw — cot A = cotB + cotC, cotw = cot A + cotB + cotC., , C, , [where k = a + b + c, given], , I Example 9. In a AABC, the tangent of half the, difference of two angles is one-third the tangent of, half the sum of the angles. Determine the ratio of the, sides opposite to the angles., Sol. We have,, , tan, , A-Bj, , 2, , 1, | A+B, = - tan, 3, 1, 2, J, , Using Napier’s analogy,, (A-B} a-b, fC, tan -------- =------ •cot —, \ 2 )• a + b, I2, , www.jeebooks.in, ■(ii)

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , From Eqs. (i) and (ii);, 1, I A + B, -tan, 3, 1, 2, , =>, , a-b, . C, ------- cot| —, a+b, 2, , 1, a-b, IC, —cot| — =, •cot —, 3, 2J a+b, I2, , as A + B + C = it, , tan, , B+C, 2, , a-b, a+b, , =>, , C, , (it, , C, , 1, 3, , 203, , or 3a - 3b = a + b, , 2a = 4b, —b —1—, or, a 2, Thus, the ratio of the sides opposite to the angles is b : a = 1:2., , = tan------- = cot—, I2 2, 2, , g Exercise for Session 1, 1. In the given figure, if AB = AC, Z.BAD = 30° and AE = AD, then x is equal to, , B, , i, (a) 15°, , C, , D, , (C)12J, , (b) 10°, , (d)7J, , 2. In AABC, a = 4, b = 12 and B = 60°, then the value of sin A is, , (a)-J,, 2V3, , (b) 3V2', , (c)-i, V3, , (d)y, , 3. Let ABC be a triangle such that ZA = 45°, ZB = 75°, then a + cf2 is equal to, , (a)0, , (b)b, , (c)2b, , (d)-b, , r3, b, 4. Angles A, B and C of a AABC are in AP. If — = -=, then ZA is equal to, c, (a)7, 6, 5. If cot—=, 2, a, (a) Isosceles, , (c)5E, , (b)£4, , 12, , (d)?, , then AABC is, (b) Equilateral, , (c) Right angled, , (d) None of these, , , a2 -b2 _sin(A-B), 6. If in a AABC, then the triangle is, sin(A + B)’, , (a) Right angled or isosceles, (c) Equilateral, , (b) Right angled and isosceles, (d) None of these, , 7 1, »■, 1, a2sin(B-C) b2sin(C-A) c2sin(A-B), * •., 111, 1 j triangle, u1 ci 1 lyiv ABC,, y ———;—- + —----- —:—- + ——-—:—- =, 7, In vii, any, sin 8 +sin C, sin C +sin A, sin A + sin B, , (a)a + b + c, , (b)a + b-c, , 8. In any AABC, if 2 cos 8 = -, then the triangle is, 2, (a) right angled, (b) equilateral, , (c) a - b + c, , (d)0, , (c) isosceles, , (d) None of these, , www.jeebooks.in

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www.jeebooks.in, 204, , Textbook of Trigonometry, , -^)(c t a-b)(a+ b-c) is equal to, , 9. The expression, , (a) cos2 A, , (b) sin2 A, , I(c) cos-A cosfi cosC, , (d) None of these, , 10. InAABC, if a cos A =b cos 8 , then the triangle is, i(b) Right angled, i(d) Right angled isosceles, , (a) Isosceles, (c) Isosceles or right angled, 11. Ina AABC, (a + b + c)(b + c -a) = Xbc, if, , (b), I X>0, (d), i X<4, , (a) X < 0, (c) 0 < X < 4, 12. If a = 9,b = 8andc = x satisfies 3 cos C = 2, then, , (a) x = 5, , i(c) x = 4, , (d)x = 7, , (c) right angled, , (d) None of these, , (b) x = 6, , 13. In AABC, if sin2 A + sin2 8 = sin2 C, then the triangle is, (a) equilateral, , (b) isosceles, , 14. The sides of a triangle are a - p, a + p and ^3a2 + P2, (a > P > 0). Its largest angle is, (b)^, 2, , (a)y, , (d)^ 5it, 6, , (c)^, , 4, , xc ., x. 1+cos(A-B)cosC, 7 5. In any tnangle,----------------- -------- =, 1 + cos(A-C)cosB, (a), , a2 + b2, aFTc2, , ' ’ b2— c2, , (c), , c2-a2, a 2 + b2, , (d) None of these, , 16. If the sides of a AABC are in AP and a is the smallest side, then cos A equals, , (b), , (a) 30 ~, , (C), 2b, , 2c, , y., , (d) None of these, , 2c, , 17. In a AABC, a2 cos 2B + b2 cos 2A + 2ab cos(A - B) =, (a) a2, , t, , (b)c2, , (c)b2, , (d)a2 + b2, , 18. In any &ABC, 2[bc cos A + ca cos B + ab cos C] =, , (b)a2+b2-c2, 1, 1, 19. Ina AABC, tan-(A + B)-cot-(A-B)is equal to, 2, (a)a2+b2+c2, , (a) a^4, +b, 20., , (b)t±*, c, , (c)a2 -b2, , (0^4, a-b, , ,2, , (d) None of these, , (d) — b, 2(a + b), , If in a AABC, b = V3, c = 1 and B - C = 90°, then ZA is, , (a) 30°, , (b) 45°, , (c) 75°, , (d) 15’, , www.jeebooks.in

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www.jeebooks.in, Session 2, Auxiliary Formulae, Trigonometric Ratios of Half-angles, , (b + c - a)(b + c + a), 2bc, , In any AABC, we have, . b2 +c 2-a 2, cos A =-------— ...(i) [using cosine law], 2bc, , A, 2sin — = l-cosA, ...(ii), 2, f b2 +c2 -a 2 \, 2sin2 —= 12, k, 2bc, , [using Eqs. (i) and (ii)], , 2bc, (a + b- c)(a - b + c), , 2 A 2sx2(s-a), 2 A, 2 cos — =---------------- ; cos, 2, 2bc, 2, , A, cos —, 2, c-, , i, , 2bc, , [we know a + b + c = 2s, =>a + b = 2s - c and a + c = 2s - b], 2sini^ = (2s~C~c)(2s~t~i’), 2, 2hc, 231,1^ =4(s-c)(s-t.), 2, 2bc, sin2 A (s - b)(s - c), 2, be, , sin A/2 is (+ve)], . A, sin— =, 2, , (s - b)(s - c), be, , (s - q)(s - c), , . C, sin — =, 2, , (s - a)(s - b), , sinA/2, tan — =, 2, cosA/2, , '($ - b)(s - c), V, , x, , be, , _ Ks ~ a^s ~c), , C, tan —, 2, , l(s - a)(s - b), , and, , be, , s(s - a), , ac, ab, , \, , s(s - b), , s(s - c), , Area of Triangle, 1, v A = - (base) (height), 2, , area of A = - BC • AD,, 2, , (A), , D A£), as sin B =-----, , = -a • (csinB), 2, •••(B), , c, , A = -ac • sinB, 2, A, , -(C), , b2 + c2- a2, A, Again, 2cos2— = 1 + cosA = 1 +, 2, 2bc, 2bc 1 b2 + c2 - a2, , 2bc, _(b + c)2~a 2, , C, s(s-c), ana cos —, 2 ” V ab, , ac, , N, , c. .. . * B, Similarly, tan —, , Similarly, it may be proved,, . B, sin— =, 2, , ls(s - b), , B, , i, , be, , If A represents the area of a triangle ABC, then, , [since in a triangle, A is always less than 180°,, , or, , s(s-a), , N, , A_ (s - b)(s -c), tan —, 2 \ s(s - a), , 2bc, , =>, , be, , cos A12 > 0., , Also,, , a2-(b- c)2, , s(s - a), , Since, A12 is less than 90°,, , Similarly, cos~, , 2bc - b2 - c2 + a2, , 2bc, q2 - (b2 + c2 - 2bc), , [where a + b + c = 2s, b + c = 2s - a], , B, , D, , AD, , kC, , www.jeebooks.in, 2bc, , Also,, , sinC =, , b, , =>AD = bsinC

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www.jeebooks.in, 206, , Textbook of Trigonometry, , Thus, area of A is -(73 + 1)., 2, , A = - a ■ b • sin C, , 2, , Similarly;, , Aliter In above example we have Z A = 105°, Z B = 30°,, ZC = 45° and a = 73 +1, , A = - be • sin A, 2, , & = — ab sin C = — be sin A = — ca sin B, 2, 2, 2, , Thus, area of, , A=, , a2 sinBsinC, 2, sin A, , l)2 sin 30°. sin 45°, , Note, (i) Area of a triangle in terms of sides (Heron's formula):, 1, 1, A, A, A = -bcsinA = -bc-2sin--cos2, 2, 2, 2, '(s - b)(s - c) ]s(s - a), = be., be, be, A = y/s(s -a)(s - b)(s - c)., (ii) Area of triangle in terms of one side and sine of three angles:, 1, 1, A = -bcsinA= -(AsinB)(/(sinC)-sinA, 2, 2, 1, , a y2, , 1, , = -/C sinAsin BsmC = •sinA-sinB-sinC, 2lsinA;, 2, sinAsinB, b2 sinAsinC, A = — sinBsinC, A, sinA, 2, sinB, 2, sinC, 2, , c, , I Example 10. If the angles of a triangle are 30° and, 45°, and the included side is (V3 + 1) cm, then prove, , sin A, , I Example 11. Consider the following statements, concerning in AABC, , (i) The sides o,b,c and area A are rational., B, Q, (ii) a, tan-, tany are rational., , (iii) o,sin A,sinB,sinC are rational., Prove that (i) => (ii) =» (iii) => (i)., Sol. a,b,c,A are rational (given)., , =>, , s, s - a, s - b, s - c are rational., , Now,, , B = (s - c)(s - a), tan —, 2, y $($ - b), , a, sin 105°, , sin 30° sin 45°, b ~, c, , b =----------- , c =, 72 sin 105°, 2sinl05°, , So, area of AABC = - be sin A = -be sin 105°, 2, , _1, , (75 + I)2, , 2 272.sin(60° + 45°), , =*, , 77, , -, , 2, , V2, , ’, , Hl.-L + _L .1, , 2(75 + 1), , A, , s2(s - b)2, , s(s - b), , B, , •, , i, , tan— = rational, [as A, s, (s - a) are rational], 2, C, Similarly, tan — is rational. Hence (i) => (ii), 2, B, 2tan —, Now, sin B =---------- — is rational by (ii)., ,B, 1 + tan —, 2, Similarly, sinC is also rational., B, C, 1 - tan — • tan —, A, B + C', 2, ■2- = rational by (ii), tan— = cot, B, C, 2, 2, tan — + tan —, 2, 2, => sin A is rational, Hence, (ii) => (iii)., .., a, b, c, Now,------ =------- =, = k, which is rational since ‘a, sin A sinB sinC?, and ‘sin A’ are rational., b, , c, and, are rational. But sin B and sin C are, sinB, sinC, rational by (iii), , .*., , 2, , ls(s - a)(s - b)(s - c), , y, , sinC, c, , sinB, ~b~, , sin (105°), , 2, , that the area of the triangle is -(V3 + 1)., Sol. We have,, , [using note (ii)], , 2, , b and c are rational., , www.jeebooks.in, 72 2, , A = - be sin A is also rational., 2, Hence (iii) => (i).

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 207, , g Exercise for Session 2, , 1. If in a AABC, (s - a) (s - b) = s (s - c), then angle C is equal to, (b) 45’, (c) 30°, B, C, A , cot —, 2. In any AABC, if cot —, , cot — are in AP, then a, b, c are in, 2, 2, 2, (a)AP, (b)GP, (c)HP, (a) 90°, , (d) None of these, , A, tan — tan 2, any AABC,-----------1- =, A, tan — + tan —, 2, 2, -, , 3. In, , (d) 60°, , (a) ^4, a+b, , X, , 6, , (b) —, c, , (c), , a-b, a+b+c, , (d), , C, A, 8, 4. In a AABC, be cos2 — + ca cos2 — + ab cos2 — is equal to, 2, 2, 2, (a)(s-a)2, (b)(s-b)2, (c)(s-c)2, , c, a+b, , (d)s2, , 5. In a AABC, ifcosA + cosC + 4sin,2: I— I, then a.b,c are in, , (a)AP, , (b)GP, , (d) None of these, , (c)HP, , C, B, 6. If in a AABC, 3a = b + c, then the value of cos — cot — is, , 2, , (b)73, , (a)1, , 7. In any AABC,, , b -c, 2 A?, COS, a, <2,, , + / c-a, , cos, , 2(b'\, (a-b\, 2( C'l, — + ------ cos — is equal to, , I 2 J, , 12J, , ,2, , (a), , a2, , (b), , (d) None of these, , (c)2, , c2 - a, b2, , UJ, , — b2, c2, , W^-A, (c), , (d)0, , 8. In a AABC, the tangent of half difference of two angles is one-third the tangent of half the sum of the two, angles. The ratio of the sides opposite the angles is, (d)3:4, (a) 2 : 3, (b) 1 : 3, (c) 2 :1, , 9., , If in a triangle, a cos2 y + c cos2, , (a)AP, , (b)GP, , = —, then its sides will be in 2, (c) HP, , (d)AGP, , 10. In the adjacent figure P’ is any interior point of the equilateral triangle ABC of side length 2 unit, , A, , ———xc, , If x„, xb and xc represent the distance of P from the sides BC, CA and AB respectively then xt + xb + xe is, equal to, , (a) 6, , <4, , (b)V3, , (d) 2j3x, , 11. Ifc2 =a2 +b2, then4s(s-a)(s-b)(s-c)isequalto, , www.jeebooks.in, (a)s4, , (b)b2c2, , 2q2, (c)c2a, 2, , (d)a2b2, , 12. The number of possible ZABC in which BC = VTl cm, CA = Vl3 cm and A = 60° is, (a) 0, , (b) 1, , (c) 2, , (d) None of these

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www.jeebooks.in, 208, , Textbook of Trigonometry, , 73. If two sides a,b and the ZA be such that two triangles are formed, then the sum of the two values of the third, side is, , (a)b2 - a2, , (b) 2b cosA, , (d)£b+c, , (c) 2b sinA, , 14. If in a AABC, sin A = sin2 B and 2cos2 A = 3 cos2 B, then the AABC is, , (a) right angled, , (b) obtuse angled, , (c) isosceles, , (d) equilateral, , (c) isosceles, , (d) isosceles or right angled, , 75. If a cos A = b cos B, then the triangle is, , (a) equilateral, , (b) right angled, , 7 6. Points D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If ZBAD = x, ZDAE = y,, i, ,sin(x + y)sin(y + z), ZEAC =z, then the value of — -----———------- is equal to, sinx sinz, (a) 1, (b) 2, (c) 4, (d) None of these, , 1°, , 1°, , 7 7. If the base angles of a triangle are 22 — and 112 —, then the height of the triangle is equal to, 2, (a) half the base, (b) the base, (c) twice the base, (d) four times the base, 78. In a AABC, a = 1 and the perimeter is six times the AM of the sines of the angles. The measure of ZA is, , (a)^, , (b)|, , (d)^, , (c)^, , 79. Ina AABC, if median AD is perpendicular to AB, then tan A + 2 tan B is equal to, , (a)1, , (b) 3, , (c)0, , 20. If p is the product of the sines of angles of a triangle, and q the product of their cosines, then tangents of the, angles are roots of the equation, (a)qx3 - px2 + (1+ q)x - p = 0, (b) px3 -qx2 + (1 + p)x - q = 0, (c) (1+ q)x3 - px2 + qx - p = 0, (d) None of these, , Session 3, Circles Connected with Triangle, , Circles Connected with Triangle, , Circum-radius, , Circumcircle of a Triangle, , The radius of the circumcircle of a AABC is called the, circum-radius given by;, a, b, c, (i)K =, (ii)B = —, 2 sin A 2sinB 2sinC, 4A, Proof, , The circle which passes through the angular points of a, AABC is called its Circumcircle. The centre of this circle is, the point of intersection of perpendicular bisectors of the, sides and called the Circumcentre. Its radius is always, denoted by R., Note, 1. Circumcentre of an a cute-angled triangle lies inside the, triangle., 2. Circumcentre of an obtuse-angled triangle lies outside the, triangle., 3. In a right angled triangle the circumcentre is the mid-point of, hypotenuse., , (i) Here, the perpendicular bisectors of the sides BC, CA, and AB intersect at 0., , .*. O is the circumcentre such that,, , OA = OB = OC = R, , www.jeebooks.in, We have,, , Z.B0C = 2^A, , ZB0D = ZC0D = ZA

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , • x bd, sin A = —, OB, , In A OBD,, , a/2, , (iii)r =, , 209, , a sin BZ2 • sin C/2, , cos A/2, b sin C/2 ■ sin A/2, csin A/2.sinB/2, r=------------------- => r =, cosB/2, cosC/2, , (iv) r = AR sin A/2 sin B/2 sin C/2, Proof Let the internal bisectors of the angles of the, triangle ABC meet at I. Suppose the circle touches the, sides BC, CA and AB at D, E and F, respectively., , Then, ID, IE, IF are perpendicular to these sides and, ID = IE - IF = r., , R = —^—, 2 sin A, , =>, , Similarly,, , R = —-— and R = —-—, 2 sin B, 2 sin C, ~b~, c, Hence, R =—-—, 2 sin A 2sinB 2sinC, , (i) We have, area of AABC = area of AIBC + area of, AL4B + area of A/CA, 1, a, 1, 1, A = - ar + - cr + - br, 2, 2, 2, , (ii) As discussed, Area of A = - be sin A, 2, a, , 2A, , sm A = —, be, Also,, , „.(i), , R = —?—, , A = -(a + b + c)r = sr, 2, , ...(ii), , A = sr, , 2sin A, , From Eqs. (i) and (ii);, _ abc, ~ 4A, , or, abc, ~ 4A, , V be;, , a +b+c, as; s =---------2, , A, r = —., s, , (ii) Since, the lengths of the tangents to a circle from a, given points are equal, therefore, , AE = AF,BD = $F and CD = CE., , In-circle or Inscribed Circle of a, Triangle, The circle that can be inscribed with in a triangle so as to, touch each of its sides is called its inscribed circle or, In-circle. The centre of this circle is the points of, intersection of bisectors of the angles of the triangle. The, radius of the circle is always denoted by lr and is equal to, the length of perpendicular from its centre to any one of, the sides of triangle., , In-radius The radius of the inscribed circle of a triangle, is called the in-radius. It is denoted by r’ and is given by, , Now,, , —(I), , 2$ = a + b + c = BC + CA + AB, , = (BD + DC) + (CE + EA) + (AF + FB), = (BD + BF) + (AE + AF) + (CD + CE), = 2(BD + AE + CD) = 2(BC + AE) = 2(a + AE), , s = a + AE, , AE=(s-a), , Now, in AIAE,, , A, r, tan —= —, 2 AE, , =>, , r = AEtan(A/2)=(s-a)tanA/2, r = (s - a) tan A Z2, , www.jeebooks.in, ..., (i) r = -, , Similarly, r = (s - b) tanB/2 and r =(s-c) tanC/2, , s, , ABC, , (ii) r = (s - a) tan — = (s - b) tan — = (s - c) tan —., 2, 2, 2, , Hence,, , r=(s-a) tanA/2, , = (s - b) tan B Z2 = (s - c) tan C !2

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www.jeebooks.in, 210, , Textbook of Trigonometry, , Formulae for rv r2, r3, , (iii) In A/BD and AICD, we have,, „ „, r, ,, C, r, tan B12 =---- and tan — =----BD, 2 CD, , r, , BD ~—-— and CD =, tan C12, tanBZ2, , Now,, , a = BD + CD, r, a =---- -—, , tanf—, I 2, , a=r, , a=r, , r, , tanf-, , cos B Z2, sin B /2, , l2, , cos C /2, sin C Z2, , cos B12 • sin C /2 + sin B12 • cos C /2, , sin B12 • sin C /2, , rsin(B/2 + CZ2), a =-------------------sin B /2 • sin C12, , In any AABC, we have, A, A, A, (i) r} = ------- > r2 =------- ’ r3 =----s-a------ s-b, s—c, (ii) r, = s tan A Z2, r2 =s tan B Z2, r3 -s tan C12, a cos B/2 ■ cos C/2, bcos A12 • cos C/2, (iii) r, =, --------------------- . r2 =, cos AZ2------------- cosB/2, _ ccos AZ2 • cosB/2, 3, cos C Z2, (iv) r, = 4F sin A Z2 cos B Z2. cos C Z2,, r2 = 4/?cos AZ2sinB/2.cosC/2,, r3 = 4R cos A Z2 cos B/2. sin C/2, , Proof (i) Let the AABC be as;, A, , [v A + B + C = 71 ], , sin(B Z2 + C/2) = sin| — - — = cos A Z2., ( 2 2, , r cos A/2, a =------------------sin B Z2 • sin C Z2, b sin A /2 • sin C /2, a sin B Z2 • sin C !2, and, r =--------------------- , r =, cosB/2, cos A/2, , c sin A/2-sin B/2, r =--------------------cos C/2, , We have,, l1D = /1B = /1F = r1, , a sin B/2-sin C/2, , _, a, andRR=, =--------(iv) We have, r = --------------------- and, cos A/2, 2sinA, , Now, area of AABC = area of Al, AC + area of Al, AB, -area of AltBC, , A = -I.E-AC + -I.F-AB--I.D-BC, 2, 2, 2, . 1 , 1, 1, A = -r,b +-r,c —r.a, 2, 2, 2, , =>, , 2R sin A • sin B /2 • sin C !2, r =-------------------------------cos A/2, , _ 2R • (2 sin A Z2 ■ cos A /2) • sin B /2 • sin C /2, , A = — (b + c-a), 2, , cos A/2, , r = 4R sin A /2 • sin B !2 • sin C Z2, , Escribed Circles of a Triangle, The circle which touches the sides BC and two sides AB, and AC produced of a triangle ABC is called the Escribed, circle opposite to the angle A., , Its radius is denoted by . Similarly, r2 and r3 denote the, radii of the escribed circles opposite to the angles B and C,, respectively. The centres of the escribed circles are called, the ex-centres., , The centre of the escribed circles opposite to the angle A, is the point of Intersection of external bisector of angles B, and C. The internal bisector also passes through the same, point. This centre is generally denoted by, , A = —(2s -2a), 2, A, r, =-----s-a, , =*, , Similarly,, , r2 =, , A, s-b, , and r3 =, , [using a + b + c = 2s], , A, s-c, , (ii) Since, the lengths of tangents to a circle from an, external points are equal,, , AE = AF, BD = BFand CD = CE, , Now, A£ + AF = (AC + CB)+(AB + BF), , www.jeebooks.in, = (AC + CD) + (AB + BD), , = AC + AB + CD + BD

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , = AC + AB + BC, , =>, , = a + b + c = 2$., , (s - a)(b - c) + (s - b)(c - a) + (s - c)(a - b), ~A, , s(b - c + c - a + a - b)-[ab ~ ac + be - ba + ac - be], , 2AB = 2s, , =>, , A, , AE = AF=s, , =>, , = - = 0 = RHS, A, b- c c - a a - b, +----- = 0, ----- +, , In AZ, AF, tan A Z2 =, , AF, , AF, , Thus,, , r2, , tan A /2 = —, s, Similarly, r2 =s tan B/2 and r3 = s tan C/2., , A, s- a, , . fB^, , tan —, , BD, D, , BD = r. tan —, 2, , =>, , =>, , Similarly, in AItCD, we have, C, CD = q tan —, , =>, , A, , cosA/2, B, C, = q tan — + tan — = q-----------------<2 2,, cos B/2cos C/2, a cos B/2 cos C/2, r*=-------------------cosA/2, , A, :------s-b, , s, , A, , s-c, , s- s + a, , s-c+s -b, , s(s - a), , (s-b)(s-c), , a, , 2s - (b + c), , s(s - a), , (s - b)(s - c), , a, s(s - a), , _, , [as, 2s = a + b + c], , a, (s - b)(s - c), , s2 -(b + c)s + be = s2 - as, , B . . C, Now, a = BC = BD + CD = q tan — + q tan —, 2, *, 2, , =>, , rt - r = r, + r3, , Sol. We have,, , (iii) In AI'BD, we have, , BD, , r3, , I Example 13. If q = r2 + r3 + r, then prove that the, triangle is right angled., , q =stanA/2, , =>, , 211, , =>, , s(- a + b + c) = be, (b + c - a)(a + b + c), be, 2, (b + c)2 - (a)1 - 2bc, , =$>, , b2 + c1 + 2be - a2 = 2bc, , =>, , bl2 +, c _, = a-2, ZA = 90°, , Similarly,, ccos A/2-cosB/2, , b cos A12 • cos C /2, , and r3 =, r2 =----------------------------------, , cos C/2, , cos B/2, , (iv) We have, q =, , a cos B/2 • cos C/2, , cosA/2, , and R = -—-—, 2sin A, , Sol. LHS, , =>, , 1, , 4Bsin A Z2 • cos A12 • cos B /2 • cos C /2, cos A Z2, , q = 4Bsin A12 • cos B Z2 • cos C Z2, , r3 =4BcosA/2-cosBZ2-sinCZ2, , ,, ., b — cc-oa-b, I Example 12. Show that------+------ +------ = 0., , G, , 2, , 2, , rcotB/2-cotC/2, , [as, r = 4RsinA/2-sinB/2-sinC/2], 4 R • sin A /2-cos B/2-cos C 72, q = RHS [as, q = 4RsinA/2-cosB/2-cosC/2], :. rcotB/2-cotC/2 = q, , I Example 15. In a right angled triangle, prove that, r+2R = s., , Similarly, r2 = 4Bcos A /2 • sin B12 • cos C/2, , G, , C, , cos B/2 cos C/2, 4RsinA/2-sinB/2-sinC/2------sinB/2 sinC/2, , 2R sin A ■ cos B /2 • cos C /2, , cosA/2, , B, , I Example 14. Prove that r cot --cot— = q., , f3, , Sol. In a right angled triangle, the circum centre lies on the, hypotenuse., (i) [-.•ZA=90°], R=2, r = (s-a)tanA/2 = (s-a)tan45°, Also,, r = (s - a), ...(u), From Eqs. (i) and (ii), we get r = $ - 2R, =>, r + 2R = s., , =>, , www.jeebooks.in, Sol. LHS, , (b-c) ! (c - a) t (a-b), q, q, q, , zl, (s “ o'! /., zl, x\ fs-aA, . .I f $ - b |, (b ~, ~T~ + <c ” aH ~T~ + (fl I A J/ ', II A /, , fs-c, ~~, t A

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www.jeebooks.in, 212, , Textbook of Trigonometry, , I Example 16. The ex-radii i\,r2,r3 of a AABC are in HP,, show that its sides a,b,c are in AP., , Solution. rp r2, r, are in HP., 2 -1, 1, =>, —=—+—, r2 5, 2(s-b) _(s-a) + ($-c), AAA, , 4k-5k-6k, , and, , 4A, , R = -^k, , A 15^7, r = — =-----s, 4, , and, , Hence, a, b, c are in AP., , I Example 17. If A, B, C are the angles of a triangle, then, prove that, , cos A + cosB + cosC = 1 + -., R, A+B, 2, , A- B, , ■cos, , But, , BD + DC = a;, , In AABD,, A+B, ., , 2, , AD, sinB, , BD, sin A / 2, ac, , sin B, , 2A, , AD - ------------------- - --------- cosec A/2, b + c sinA/2 b + c, A,, Al, Also, —, , ID, , AB, , c, , b+c, , BD, , ac, , a, , (ii), , [using Eq. (i)], , b+c, , [as, r = 4jRsinA/2-sinB/2-sinC/2], r, cos A + cosB + cosC = 1 + —, , R, , I Example 18. Find the ratio of the circum-radius and, the inradius of AABC, whose sides are in the ratio, 4:5:6., a = 4k, b = 5k, c = 6k, , 15k, s ------, , —(i), , 2, , A = Js(s - a)(s - b)(s - c), , or, , ID, , a, , Al, , b+c, , On adding T’, we get, ®+l. —^— + 1 => ID + Al, Al, b+c, Al, , a+b+c, b+c, , =>, , a+b+c, , Al =--+ c------ 2A, cosec A/2 = — ■ cosec A/2, a+b+c b+c, s, Similarly,, , l*-5k, 2, , C, , D, , BD = ——a, b+c, , n . C n . A . B, = 1 + 2sm — -2sin—sin —, 22 2, . . A . B . C, = 1 + 4 sm—sin — -sin — = 1 + —, 2, 2, 2, R, , 2, , [using Eqs. (iii) and (iv)], , BD: DC = c : b, , B, , - = 90° 2, , 2, , 7, , yf/k/2, , I Example 19. Find the ratio of IA: IB: IC, where I is the, incentre of AABC., , 2, , 15k (15k, --------------- 4k, , r, , + cosC, , 2, , C, (A + B, A- BA, - cos, = 1 + 2sin — cos, 2, 2, 2, , Sol. Here,, , 16, , R:r = 16:7, , =>, , A-B, ,C, n . C, = 2sin —-cos, + 1 - 2sin —, 2, 2, 2, C, = 1 + 2sin— cos, 2, , (iv), , 2, R _8k/^7, , Sol. Here,, , Sol. cos A + cosB + cosC, , = 2 cos, , [using Eqs. (i) and (ii)], , 15k, , r~—-k, , 2b = a + c, , =>, , .(iii), , V7, , 2s - 2b = 2s - (a + c), , =>, , [using Eq. (ii)), , 4-^k‘, 4, , BI =—■ cosec B/2, s, , CI = — cosec C/2, s, , 2, , www.jeebooks.in, 15a/7, , k2, , 4, , ...(ii), , => IA:IB:IC = —■ cosec A12: — cosec B/2-. —cosec C/2, s ., s, s, :. IA .IB:IC = cosec A/2: cosec B/2: cosec C/2

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www.jeebooks.in, 214, , Textbook of Trigonometry, , 10. In any AABC, show that cos A + cos B + cos C = I 1 + — |., R, 11. If the side be a, b and c, then show that, , a, , — -r, ~b~', , £, r3, , 12. Show that r2r3 + r3r\ + r/2 = s2, 13. Show that (r, + r2 )(r2 + r3 )(r3 + r,)=4Rs.2, 14. If g = r2 + r3 + r, then show that A is right angled., , 15. In an equilateral triangle, show that the in-radius, circumradius and one of the ex-radii are in the ratio 1:2:3., 16. Show that, , 1, r, , 1, , 1, r, , 1, , 1, r, , 1, , 16R, r2(Za)2, , 17. If^,r2,r3 in a triangle be in HP, then show that the sides are in AP., , 18. In a AA8C, show that r2 r3 = A2., 19. If/,, /2, /3 are respectively the perpendicular from the vertices of a triangle on the opposite side, then show that, , 1 2 3, , 8R, , 20. If the angle of a triangle are in the ratio 1:2:3, then show that the sides opposite to the respective angle are in, the ratio 1: V3 :2., , >4, , B, , C, , 21. Show that, 4 Rr cos — cos — cos — = S, 2, 2, 2, 22. If (a -b)(s -c) = (b -c)(s - a), then show that r„r2,r3 are in HP., , 23., To show, show that, that -r4’ +, + 4’ + 4’ + 4'== ~2~, 23. To, r2 r2 r2 r2, S2, 24. Show that (r, -r)(r2 -r)(r3 -r)=4Rr.2, , 1Y J. i, , 1Y1, 1, 25. Show that - + — — + —, A G. lf2 GAG, , A, , 64R3, a2b2c2, , 26. If the sides be a, b and c, then find the value of (r + r,) tan, , B-C, , 2, b — c c-a a -b, 27. If the sides bea,b,c, then find value of, +------ +------ ., r,, r2, 28. If the sides be a, b, c, then find, , + (r + r2) tan, , A-B, C-A, + (r + r3)tan, 2, 2, , - r )(r2 + r3)., , 29. If a,b,c are in AP, then show thatrvr2,r3 are in HP., , 30. Show that ^ + ^-+-^- = 2R-r, be ca ab, 31. Show that rA + r2, , -cco(g., , 32. Show that Rr(sin A + sin 8 + sinC) = A, , 33. Show that 16R2r, , r2 r3=a2b 2c2, , www.jeebooks.in, 34. If-= —, then show that c =90°., r3

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www.jeebooks.in, Session 4, Orthocentre and Its Distance from the Angular, Points of a Triangle and Pedal Triangle and, Centroid of Triangle, Orthocentre and Its Distance, from the Angular Points of a, Triangle and Pedal Triangle, Let ABC be any triangle, and let AK, BL and CM be the, perpendicular from A, B and C upon the opposite sides of, the triangle. These three perpendiculars meet at a point O', which is called the orthocentre of the triangle ABC. The, triangle KLM, formed by joining the feet of these, perpendiculars is called the pedal triangle of ABC., A, , /, /, , • Some Relations between Orthocentre,, Incentre, Escribed Circles, Centroid,, Circum-centre and Pedal Triangle, , Pedal, , mA-, , '9^-C \, B, , =>, , (i) Orthocentre of the triangle is the incentre of the pedal, triangle., , c, , K, , InAO'BK, tan(90°-C) =, , O'K, , KB, O' K = KB. tan(90° - C) = KB cot C, DIZ, , = AB cos B cot C V from AABK, cos B = —, AB, „ cosC, = c•cos B------sinC, , => O'K = 2RcosBcosC \’R = —-—, 2 sin A, , Similarly, BO' = 2R cos B and CO' = 2R cos C, Thus, the distance of the orthocentre of the triangle from, the angular points are,, AO' =21? cos A, BO' = 2R cos B, CO' = 2R cos C,, and its distance from the sides are,, O' K = 21? cos B cos C., O' L=2RcosCcosA., O' M = 21? cos A cos B., , b, , c, , 2sinB, , 2sinC, , Similarly,, , O' L = 2R cos A cos C and O' M = 2RcosAcosB, AL, InAAO'L, cos(90° - C) = ——, ' AO', AO' = AL • cosec C => AO' = ccos A • cosec C, •: from AALB, cos A = —, AB, , (ii) If 1,, I2 and I3 be the centres of escribed circles which, are opposite to A, B and C respectively and I is the, centre of incircle then AABC is the pedal triangle of, the A/tf2l3 and I is the orthocentre of the AIJ2Ir, , (iii) The centroid of the triangle lies on the line joining the, circumcentre to the orthocentre and divides it in the, ratio 1 : 2., , (iv) Circle circumscribing the pedal triangle of a given, triangle bisects the sides of the given triangle and also, the lines joining the vertices of the given triangle to, the orthocentre of the given triangle. This circle is, known as nine point circle., (v) Circum-centre of the pedal triangle of a given triangle, bisects the line joining the circumcentre of the, triangle to the orthocentre., , I Example 23. In AABC, a,b and c represents the sides,, thus find the sides and angles of the pedal triangle., , www.jeebooks.in, =>, , AO' = c- cos A——, sinC, , =>, , AO' =2R cos A, , Sol. Let AABC be any triangle and let D, E, F be the feet of, perpendicular from the angular points on the opposite, sides of the AABC, then the ADEF is known as Pedal, triangle of ABC.

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www.jeebooks.in, 220, , Textbook of Trigonometry, , < a ■ B . C, I B+C, = 1 - 8sm — -sin — ■ cos, 2, 2, 1, 2, , o . B . C . A, = 1 - 8sin—sin—sin—, 2, 2, 2, 01 ir——:---------------— = Jl - 8sin B/2-sinC/2-sin A/2, R y, , = R2 + 16B2sin2B/2-sin2C/2-, , 8R2 sin B/2sin C12 cos, , C-B, 2, , C, ^ = l + 16sin2--sin2-R, 2, 2, C, D . B . C,, „, B, -C, . B, ~, 8sin—sin—| cos —cos—Fsin — "Sin —, 2, 2, 2, 2, 2, 2, B, C(, B, C, B, C, = 1 - 8sin —-sin — cos —-cos — - sin—sin—, 2, 2, 2, 2, 2, 2, , or, , 01 =RJ1- —, V, R, 01 = ^R2 - 2Rr., , Exercise for Session 4, 1. If H is the orthocentre of the AABC, then find AH., 2. A circle touches two of the smaller sides of a AABC (a<b < c) and has its centre on the greatest side. Then,, , find the radius of the circles., , 3. If the sides be a,b,c, then show that a cos A + b cosB + ccosC = 4Rsin>4sinBsinC, 4. If the altitudes of a triangle be 3, 4, 6, then find its in-radius., , 5. In a AABC, if a =3, b =4, c = 5, then find the distance between its incentre and circumcentre., 6. If pv p2, p3 are respectively the perpendicular from the vertices of a triangle to the opposite sides, then find the, , value of p1p2p3., , 7. Show that the distance between the circumcentre and the incentre of the triangle ABC is ^R2 -2Rr., , 8. Show that the distance between the circumcentre and the orthocentre of a triangle ABC is, R ^1 - 8 cos A cos 8 cos C., , 9. If in a AABC, AD, BE and CF are the altitudes and R is the circumradius, then find the radius of the DEF., 10. If/,lvl2 and /3 be respectively the centre of the in-circle and the three escribed circles of a AABC, then find /2/3., , www.jeebooks.in

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www.jeebooks.in, Session 5, *—MXKKWirwta— ',7' > ■", , »« *«-!, , ~ .---TV, , ~>g, , trx-» M -*3C- mi n, , ■ au, , na Ml II, , BBzlV-’j* _■«* -z—: «., , Regular Polygons and Radii of the Inscribed, and Circumscribing Circle a Regular Polygon, _2, , it, , ., , 7i, , A regular polygon is a polygon which has all its sides as, well as all its angle equal., , = nR -cos —-sin—, n, n, , If the polygon has ‘n sides, Sum of the internal angles is, , n 2 ■ 2n, = — R sin —, 2, n, , (n — 2)n and each angle is ——, n, A, o, , D, , \R', , /, < /, , /, , ', , 360°, , jHI-V, , Sol. We know, in hexagon central angle is —— - 60° and each, angle =, , *, , a, , I Example 32. If Ao, A,, A2,A3, A4 and A5 be the, consecutive vertices of a regular hexagon inscribed in a, unit circle. Then, find the product of length of, A0A,,A0A2 and A0Afl., , Let AB, BC and CD be three consecutive sides of the, regular polygon and n be the number of its sides. Let 0 be, the point of intersection of the bisector of the angles, ZABC and ZBCD., , (2n - 4)n, 2n, (6 - 2) x 180°, 6, , = 120°, , The point 0 is both the incentre and circumcentre of, polygon and so BL = LC. Hence we have,, A2, , OB = OC = R, the radius of the circumcircle and OL = r, the, radius of the incircle., It can be seen,, D, , a, , R = —cosec, , 2, , ,, a /71, and r =—cot —, 2, , I*., , where ‘a is length of a side of the polygon., The area of the polygon = n area of AOBC, a, fit} a, a, = n(OL)(BL) = n — cot — •, 2, nJ 2, , As the unit circle,, /., radius OA0 = 1 = r, In AAOA}AV, A0A,2 + AxA22 - A0A2, cos 120° =, 2AoA1-A1A2, , 1 + 1 - aoa2, 2, 2-1-1, , / 71?, 1, 2, - — na •cot —, 4, , =>, , AoA2 = V3, , (i), , Similarly in AA0A5A<, we have, A0A4=^, , Also, the area of the polygon., = n(OL)(BL) = n(OL)(OL tan ZBOL), , 2 . f 71, = nr 2 tan —, , ...(ii), , Thus, the value of,, (A0X,)-(A,A,)-(A0A,) = l-^-5/3, , = 3 square units, , www.jeebooks.in, Again, the area = n (OL)(BL)., , 71, , = n OB • cos — - OB • sin —, n, n, , I Example 33. If the area of circle is A, and area of, regular pentagon inscribed in the circle is A2, then find, the ratio of area of two.

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www.jeebooks.in, =, , 223, , Chap 03 Properties and Solutions of Triangles, , Exercise for Session 5, , 1. Find the sum of the radii of the circles, which are respectively inscribed and circumscribed about a regular, polygon ofn sides., 2. Find the radius of the circumscribing circle of a regular polygon ofn sides each of length is a., , 1111, 3. If A A ■ A,, A be the area of the in-circle and ex-circles, the show that -= + -= + -7= = -7=., VA VA VA VA, , 4. A regular polygon of nine sides, each of length 2, is inscribed in a circle, then find the radius of the circle., 5. Show that the area of the circle and the regular polygon of n-sides and of equal perimeter are in the ratio of, 71 l n, tan I —, :, \nj n, , 1, , 1, , 1, , 6. Let A, A> A> •■■ ■ A be the vertices of an n-sided regular polygon such that----- =------ +------ . Find the value, , AA AA, , AA, , of n. Prove or disprove the converse of this result., , 7. Let ln is the area of n-sided regular polygon inscribed in a circle of unit radius and O„ be the area of the polygon, _____, , ■I, , ', , If, , ___ ■, , _•, , I, , “T"l. _, , _ . ., , £ i— _ X, , 21,, , f, , 2, , circumscribing the given circle. Then, prove that ln = — 1 + J1- —, 2, V V n ., 7, \, , Session 6, esxwwmw wi, , i'1Bi M, , mfiiww’WTi n1, , "c-, , jj, , i'»ui .*r»» —iHiiii n, , Quadrilaterals and Cyclic Quadrilaterals, Area of Quadrilateral, , Similarly,, , ABCD is any quadrilateral where AB = a, BC = b, CD = c,, AD = d and Z.DPA = a., Let $ denotes the area of quadrilateral, then, area of ADAC = area of AAPD + area of ADPC, , a, , ...(ii), , c, , = -DP • AC • sina + - BP • AC • sina, 2, 2, [using Eqs. (i) and (ii)], , b, , = -(DP + BP)- AC -sin a, 2, , D, , A, , area of AABC =-BP ■ AC • sina, 2, s = area of ADAC + area of AABC, , s=-BD- AC-sina, 2, , B, , Area of quadrilateral = - (product of the diagonals), 2, x (sine of included angle)., , 1, 1, = -DP • AP • sina + - • DP • PC • sinfn - a), 2, 2, , www.jeebooks.in, = if)P.(AP + PC)sina, , Area of ADAC = - DP -AC - sina, ., 2, , (i), , Again,, We can express the area of A in terms of sides and the sum, of two opposite angles :

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www.jeebooks.in, 225, , Chap 03 Properties and Solutions of Triangles, , Ptolemy's Theorem, , ABCD = R = Circum-radius of AABC =, , In a cyclic quadrilateral ABCD,, , AC(ab + cd), , AC-BD = AB-CD + BC-AD, But, AC, , Proof: Let ABCD be a cyclic quadrilateral, where, , Hence,, , l(ac + bd)(ad + be), y, , 2sinB, , [using A = -(ab + cd) sin B], 2, , 4A, , i.e. in a cyclic quadrilateral the product of diagonals is, equal to the sum of the products of the lengths of the, opposite sides., , AC, , (ab + cd), , [using Eq. (i)], , R = — 7 (ac + bd)(ad + bc)(ab + cd), 4A, R, , 1 I (ac + bd)(ad + bc)(ab + cd), 4, , (s - a)(s - b)(s - c)(s - d), , I Example 36. If the sides of a cyclic quadrilateral are 3,, 3, 4, 4. Show that a circle can be inscribed in it., „2, , and, , cosB =, , ,, , 2, , >2, , i2, , a +b —c —a, , 2(ab + cd), =>, , AC2 =a2 + b2 -lab­, , a2+b2-c 2-d2, , 2(ab + cd), =>, , . „2, , c, , Sol. By geometry,, , AC2 =a2 +b2 - labcosB, , (a2 + b2) (ab + cd) - ab(a2 +b2 -c 2-d2), (ab + cd), , .rz _(a2 + b2)- cd + ab(c2 +d2), ab +cd, , (ac + bd) • (ad + be), AC-------------------------ab + cd, , ...(0, , Note, , Similarly,, , BD2 _(ab + bd) • (ac + bd), ad + be, , AP = AS, ...(i), ••■(ii), BP= BQ, f, R, \, t\, DR = DS, ...(iii), ...(iv), CR = CQ, ?Q, s', Adding all four equations, we get, IB, AB + CD = AD + CB, ...(v), A, Now, the sides of cyclic quadrilateral, 3, 3, 4,4 inscribe the circle in it, if it, satisfy (v)., i.e., let AB +CD = 3 + 4= 7, AD + CB = 3+ 4 = 7, i.e., 3,3, 4,4 i.e. 3 = AB, 3 = BC, 4 = CD, 4 = CB, satisfy condition (v) or a circle can be inscribed., , If sum of opposite side of a quadrilateral is equal, then and only, then a circle can be inscribed in the quadrilateral., , ...(ii), , =>, , AC2 - BD2 =(ac + bd)2, , =», , AC • BD = (ac + bd), , =>, , AC - BD = AB -CD + BC - AD [ Ptolemy’s theorem], , Circum-radius of a Cyclic Quadrilateral, Let ABCD be a cyclic quadrilateral. Then the circum-circle, of the quadrilateral ABCD is also the circum circle of, AABC., , I Example 37. The two adjacent sides of a cyclic, quadrilateral are 2 and 5 the angle between them is, 60°. If the area of the quadrilateral is 4 Vs, then find, the remaining two sides., B, , So/. Let AD = 2, AB = 5, ZDAB = 60°, , Since, the quadrilateral is cyclic,, ZBCD = 120°, Area of, , 1, -Ji, AABD = -2-5sin60°=5.—, , ...(i), 2, 2, Area of ABCD = Area of, quadrilateral ABCD - area of AABD, , 2, , y XX, 120*', , 5,, , Iw, , X, , 2, , D, , 2, , www.jeebooks.in, ...(ii) [using Eq. (i)], , Let CD = x and BC = y, , Hence, the circum-radius of the cyclic quadrilateral, , Now, area of ABCD = xysin 120°

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www.jeebooks.in, 226, , Textbook of Trigonometry, , 3yf3 1 73, ---- = -.Vy---2-- 2, 2, =», xy = 6, Applying cosine rule in ABAD, we get, , or, , I Example 38. If a,b,c,d are the sides of a quadrilateral,, , •(Hi) [using Eq. (ii)], , then find the minimum value of, , (iv), , =>, , BD2 = 19, , 1, 2, , x2 + y2 + xy = 19, , or, , x2+y2 = 13, , B, , And we know, (a - b)2 + (b - c)2 + (c - a)2 > 0, 2(a2 + b2 + c2) > 2[ab + be + ca), 3(a2 + b2 + c2) > (a2 + b2 + c2) + 2(ab + be + a), [i.e., adding (a2 + b2 + c2) to both sides], , [using xy = 6], , 3(a2 + b2 + c2)>(a + b + c)2, , x2 + y2 + 2xy = 13 + 12 = 25, , (x + y)2 = 25, =>, and, , a, , A, , x2 +y2 - 19, 2xy, , or, Now,, , ', , 1 _ 22 + 52 - BD2, 2~, 2-2-5, , Applying cosine rule in ABCD, we get, x2 +y2 - 19, cos 120° =------ - -------2xy, or, , d1, , Sol. Here, AB = a, BC = b, CD = c and AD = d are the sides of, quadrilateral ABCD., , AD2 + AB2 - BD2, cos60° =, 2AD-AB, or, , Q2 + b2 + C2, , =>, , [v sum of any three sides, of quadrilateral is greater than fourth], 3(a2 + b2 + c2) > (a + b + c)2 > d2, [v a + b + c > d], , x+y=5, x2 + y2 -2xy = 13-12, , a2 + b2 + c2 1, d2, >3, , x-y=±1, Solving we get, (x = 3, y = 2) or (x = 2, y = 3), , Minimum value of, , a2 + b2 + c, , d2, , \1., 3, , Exercise for Session 6, 1. The area of a cyclic quadrilateral ABCD is, , (3V3), 4, , ■■ The radius of the circle circumscribing cyclic quadrilateral is 1., , If AB = \ BD = 73, then find BC • CD., , 2. If two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60°. If the third side is, 3, then find the remaining fourth side., 3. The ratio of the area of a regular polygon of n sides inscribed in a circle to that of the polygon of same number, of sides circumscribing the same circle is 3 :4. Then, the value of n is, 4. A right angled trapezium is circumscribed about a circle. Find the radius of the circle. If the lengths of the bases, (i.e. parallel sides) are equal to a and b., , 5. If A B, C, D are the angles of quadrilateral, then find, , Ztan A, EcotA’, , www.jeebooks.in

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www.jeebooks.in, Session 7, Solution of Triangles, i, , i, , In a triangle, there are six variables, viz. three sides a, b, c, and three angles A, B, C when any three of these six, variables (except all the three angles) of a triangles are, given, the triangle is known completely; that is the other, three variables can be expressed in terms of the given, variables and can be evaluated. This process is called the, solution of triangles., , Case II When two sides a, b and included ZC are given., In this case we use the following table :, ____, , Given__________________ Required_________, , o, band ZC, , (ii) tan|, , Case I When two sides are given., , Required, , Given, (>) a, b, , tan A = -, B = 90° - A, C = —, sin A, , £, , (ii) a,c, , sin A = -, b = ccosA, B = 9(f - A, c, where ZC = 90°, , Case II When a side and an acute angle is given. In this, case we can determine the remaining variables as given in, the following table :, ____ Given, , Required, , (i) a,A~, , a, B = 9(F - A, b = acot A, c =----_____sin A, B = 9(P - A, a = csin/t, b = ccosA, , (ii) c,A, , Solution of a Triangle in General, Case I When three sides a, b and c are given., , In this case the remaining variables are determined by, using the following table :, , Required, , Given, , a, b, c, , (i) Area of A = ^5(5 - a)(s - b)(s - c),, 2s = a + b + c, . J 2A ., 2A . _ 2A, (ii) sin A =—, sinB =—, sinC =—, v}, be, ac, ab, A, A, BA, tan — =-------- , tan — =-------- ,, 2 5(5 - a), 2 s(s - b), C, A, tan — =------2 s(s - c), , ' A- B, < 2, , a+ b, , V2), , asinC, (iii) c =------sin .4, , Solution of a Right Angled Triangle, Let the triangle be right angled at C, then we can, determine the remaining variables as given in the, following table:, , (i) Area of A =^absinC, , Case III When one side a and two angles A and B are, given. In this case we use the following table :, Given, , a and Z/i, ZB, , Required, (i) ZC = 180? - (Z4 + ZB), asinB , asinC, (ii) b = ------ and c=------sin A, sin A, (iii) A = ^acsinB, , Case IV When two sides a, b and Z.A opposite to one side, is given then,, . r. b . ., sm B = — sm A, a, ZC = 180°-(ZA +ZB); c = ——, sin A, , From Eq. (i), the following possibilities will arise :, (a) When A is an acute angle and a < b sin A In this, , relation sin B = - sin A gives that sin B > 1 which is, a, impossible., Hence, no triangle is possible., (b) When A is an acute angle and a = b sin A In this, case only one triangle is possible which is right, angled at B., , a = bsinA => ZB =90°, (c) When A is an acute angle and a > b sin A In this, , case there are two values of B given by sin B = —a, say Bj and B, such that B, + B, = 180° side ‘c’ can be, , www.jeebooks.in, (iii), , obtained by using c = -■, sin A, , a > b sin A => two triangle are possible.

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 229, , - 5, b 4, , 5. Ina AABC, if r, = 2r2 = 3r3, then show that - =, , *, , ', , 6. Find in-radius of the triangle formed by the axes and the line 4x + 3y -12 = 0., , 7, Ina &PQR as shown in figure given that x:y:z::2:3:6, then find value of Z.QPR., , D, , 8. In a AA8C, if — < 2, then show that the triangle is equilateral., r, 9. The angle of a right-angled triangle are in AP. Then, find the ratio of the in-radius and the perimeter., , w, , A, , 10. If in a triangle 1-— 1-— = 2, then show that the triangle is right angled., r2 JI r3 >, , Session 8, Height and Distance, Angle of Elevation, If‘0’ be the observer’s eye and OX be the horizontal line, through 0. If object P is at a higher level than eye, then, /.POX is called the angle of elevation., , Note, (i) Angle of elevation and depression are always acute angle., (ii) Angle of elevation of an object from an observer is same as, angle 0 depression of an observer from the object., , Bearing, o*7, , 7/, o', , L0_________, Horizontal line, , X, , If the observer and the object are 0 and P be on the same, level respectively, then bearings is defined. To measure, the bearing the four standard direction East, West, North, and South are taken as the cardinal directions. Angle, between the line of observation, i.e., OP and any one, standard direction is measured., N, , Angle of Depression, , p, , If ‘O’ be the observer’s eye and OX is a horizontal line, object P is at a lower level than 0, then the Z.P0X is called, the angle of depression., Horizontal line, , ■x, , www.jeebooks.in, s, , p, , Thus, XPOE is called the bearing of the point P with, respect to 0 measured from East to North.

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www.jeebooks.in, 230, , Textbook of Trigonometry, , In other words the bearing of P as seen from O is the, direction in which P is seen from O., , Sol. (a), , P, , Note, North-East means equally inclined to North and East. ENE, means equally inclined to East and North-East., , I Example 42. Two flagstaffs stand on a horizontal, plane. A and B are two points on the line joining their, feet and between them. The angles of elevation of the, tops of the flagstaffs as seen from A are 30° and 60°, and as seen from 8 are 60° and 45°. If AB is 30 m, the, distance between the flagstaffs in metres is, (a) 30 + 1571, (b) 45 + 15>/3, (c)60-15a/3, (d) 60+ 1573, Sol. (d) Let x and y be the heights of the flagstaffs at P and Q, respectively., AQ = ycot30° = yJ5, , Then, AP = xcot60° =, V3, , R, , ° G, D., H, , A, , = 52 + 52 = 50, , Also, Z.THG = 90°,, (TU)2 =(TH)2 + (GH)2, = 50 + 25 = 75, Let 9 be the required angle of elevation of G at T., n GH, Then,, sin 9 =-----, , V3, , BP- AP = x -= AB, V3, , TG, 5, , 1, , 573, , 73, , 9 = sin, S, , x, y, , __ 60°f, t45°___, Q, B, 30 m, , 30', , .A60°, A, , P, , B, , T, , Let H be the mid-point of BC since, ZTBH = 90°,, (TH)2 = (BT)2 + (BH)2, , BP = x cot 45° = x, BQ = ycot60° = -7=, , R, , C, , Ji), , I Example 44. Each side of an equilateral triangle, subtends an angle of 60° at the top of a tower h m, high located at the centre of the triangle. If a is the, length of each side of the triangle, then, (a) 3a2 = 2h2, (b) 2o2 =3h2, (b) o2 = 3h2, (d)3o2 = h2, Sol. (b), , 3o73 = (73 - l)x, 60°, , x = 15(3 + 73), , Similarly,, , So that,, , 30 = y, , => y = 15^3, , V3., , PQ = BP + BQ = x + ^=, = 15(3 + 713) + 15 = (60 + 15^3) m, , I Example 43. In a cubical hall ABC D, P Q R S with, each side 10 m, G is the centre of the wall B C R Q and, T is the mid-point of the side AB. The angle of, elevation of G at the point T is, , (a) sin, , <71., , (b) cos, , _1_, , <71, , Let O be the centre of the equilateral triangle ABC and OP, the tower of height h. Then, each of the triangles PAB, PBC, and PCA are equilateral., Thus, PA = PB = PC = a., Therefore, from right-angled triangle POA, we have, • PA2 = PO2 + OA2., , www.jeebooks.in, (c) tan, , 7j_, , 173, , (d) cot, , 1, , <73

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 2, , a2 = h2 + I -sec30°, , 231, , Sol. (d) AP= PB = hcot30P = 73h, , 3h2 4- 3h2 - 602, , 4 3, , ,2 —, a2, = h+, 3, , =>, =>, , 2 2, , ,2, , -a = h, 3, 2a2 = 3h2, , I Example 45. A vertical tower PQ subtends the same, angle 30° at each of two places A and B, 60 m apart, on the ground, AB subtends an angle 120° at the foot, of the tower. If h is the height of the tower, then, 9h2 + h+l is equal to, (a) 3121, (b) 2136, (c) 3600, (d) None of these, , - = cos 120° =, 2, 2 3h2, -3h2 = 3/i2 +3/12 -3600, 9h2 = 3600, , h = 20, 9h2 + h + 1 = 3600 + 20 +1, = 3681, O, , HP, ,30°, , A, , 120°, , 60m, , B, , Exercise for Session 8, 1. If a tower subtends angles 0,20 and 30 at three points A B and C respectively, lying on the same side of a, horizontal line through the foot of the tower, show that — —, —2.e., BC cot20-cot30, 2. A person stands at a point A due south of a tower of height h and observes that its elevation is 60°. He then, walks westwards towards B, where the elevation is 45°. At a point C on AB produced, show that if he find it to, be 30°. OA, OB, OC are in GP., 3. A train travelling on one of two intersecting railway lines, subtends at a certain station on the other line, an, angle a when the front of the carriage reaches the junction and an angle 0 when the end of the carriage, reaches it. Then, the two lines are inclined to each other at an angle 6, show that 2 cot 0 = cot a - cot 0,, cota + cotp, , 4. The angle of elevation of the top of the tower observed from each of the three points A S, C on the ground,, forming a triangle is the same angle a. If R is the circum-radius of the triangle ABC, then find the height of the, tower R tan a., 5. The length of the shadow of a pole inclined at 10° to the vertical towards the sun is 2.05 metres, when the, elevation of the sun is 38°. Then, find the length of the pole., , www.jeebooks.in

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www.jeebooks.in, JEE Type Solved Examples:, Single Option Correct Type Questions, c Ex. 1. //tan, , 2 71 - B, , 2 71 - A, , + tan ----- + tan, , 2 n-C, , 4, , V2 _f2, , n-B, , n-C, , 4, , 4, , ~, , B, , H, , M 1/V3 H, a, , • Ex. 4. In tsABC, if AC =8, BC =7 and D lies between A, [From Eq. (i)], , and B such that AD = 2, BD = 4, then the length CD equals, (a) /46, (b) V48, (c) ^51, (d) ^75, , Sol. (c) I2 = 22 + 82 - 2-2-8cosA, , equals to, , = 4 +64 -32 cos A, 62 + 82 - 72, and cos A =, 2-6-8, , cot A +cotC, , (b)^, , A, 2/\, , cotB, , (a)^, , dL, (d)—, (c) —, 2001, 2001, cot A + cotC _ sin(A + C)sinB, Sol. (b), cotB, sin A sin C sin B, , B, , sin2B, , 4K2b2, , sin A cos B sinC, , 4R2 ac cosB, , 2b2, , 2b.2, , 2ac cosB, , a2+c2-b2, , 2, 2002 b2 - b2 ~ 2001, , 51, , 17, , 16X6, , 32, , \8, I \, , 4, , 2b2, , 17, , r = 68-32 X —— = 51, 32, , 1 = V51, , • Ex. 5. In a triangle, if, (a + b + c)(a + b-c)(b+ c -a), a2+b2+c2’, , • Ex. 3. A triangle has vertices A, B and C and the respec­, , tive opposite.sides have lengths a, b and c. This triangle is, inscribed in a circle of radius R. Ifb-c-l and the altitude, from A to side BC has length, , [2, , then R equals, , C, , 7, , 8a2b2c2, , (c)T, , 1/V3, , 2y[2, , • Ex. 2. IntsABC.a2 + c2 = 2002b2,, , (a)-^, V3, , A., 1, , 3, , 5/3, , 71, , A=B=C, , then, , 2, , _ (2)0)0), , a =p = y, n-A, , A, , 4A, , (i), , a +p+y =—, 2, Etana -tanfJ = 1, Etan2a = 1 = Etana tan0, tana = tan(J = tany, , 4, , 1, , and A = - ah, <3, 2, 2 5/3, , —Y=—, n, , =>, , 2, , _ 1, , (b) isosceles, (d) None of these, n-B, n-C, , Sol. (a) Let a =, , =>, , Sol. (d) a =, , 4, , 4, , &ABC is, (a) equilateral, (c) scalene, , = 1, then, , (a) isosceles, (c) equilateral, Sol. (b) We have,, , (b)-4, , then the triangle is, , (b) right angled, (d) obtuse angled, , s(s - a)($ - b)(s- c) =, a2+b2+c2 =, , V3, , As,, , A = -besin A =, 2, R, , abc, , 2A2, , a2b2c2, , 2(a2 +b2 + c2), , ...(0, , www.jeebooks.in, 2V2

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , So, Eq. (i) becomes, , • Ex. 8. In &ABC, if A A = 30°, 5 = 10 and a = x, then the, , (a2b2c2), , values ofx for which there are 2 possible triangles is given, by (All symbol used have usual meaning in a triangle), , SR2, 2a2-b2-c2, =>, 4P2(sin2A + sin2 B + sin2C) = 8S2, =>, sin2 A + sin2 B + sin2 C = 2, or, 2 + 2cosA-cosB-cosC =2, =>, cos A • cos B • cos C = 0, c. AABC must be right angled., a2 + b2 + c2 =, , I, , 233, , (a)5<x<10, , (b)x<2, , (c)-<x<10, 3, , (d)-<x<10, 2, , Sol. (a) If c is the third side, then the altitude to c has length, 10sin30° = 5,, , • Ex. 6. Consider a tsABC and let a, b and c denote the, , there are two triangles if x is greater than this value and, less than the length of b., Ifx > 10, pointB will come to the left of A and ZA would be, obtuse in the case., , lengths of the sides opposite to vertices A, B and C, respec­, tively. Ifa = 1, b = 3 and C = 60°, then sin2 B is equal to, , J, , c, , (c)S, , (d)7, , So/, (a) By Cosine law, cos 60° =, , l2 +32 -c2, , A, , 2-1-3, , c = 77, , =>, , b, XT, c, Now,------ =, sinB sinC, bsinC, sinB =, c, , [By Sine law], , A, , ____ 2_, , 77, _ 3V3, 8, , 277, • 2n, , Hence,, , 27, , sin B = —., 28, , • Ex. 7. In CsABC, if cos A + sin A -, , A, G, , c, , 2, , ., , ------ is equal to, c, , (a)Vz, , (b)1, , (c)T, V2, , (d) 2V2, , Sol. (a) We have,, , and, =>, and, =>, , s;, , 8, , 82, , V3, Aliter We have, — = cos 30°, 2, _ 100 + c2 - x.2, - [by using cosine rule], 2(10)(c), => c2 - io73c + (100 - x2) = 0, , 1073 ± ^4x2 - 100, c=, , 2, , Now, for 2 distinct positive values of c, we must have, 1077 > ^x2 - 100, , =>, =>, , 300 > 4x2 -100, x2 < 100 => 5 < x < 10, , and AB respectively such that CB = BP = PQ = QA. If, Z.AQP = B, then tan2 6 is a root of the equation, (a) y3 + 21y2 - 35y - 12 = 0, (b) y3 -21y2 + 35y- 12 = 0, (c) y3 -21y2 + 35y-7 = 0, (d) 12y3 -35y2 +35y-12 = 0, Q, , => cos A cos B + sin A sin B + cos A sin B + sin A cos B = 2, =>, , 30°, , • Ex. 9. In a tsABC, AB = AC, P and Q are points on AC, = 0, then, , cosB + sin B, , a+b ., , x=a=5, , /"x=5, , 3, , 2o, , cos(A - B) + sin(A + B) = 2, , cos(A - B) = 1, sin(A + B) = 1, A = B, so a = b, sin 2 A = 1, A = 45°, A = 135° (Not possible), a + b _ 2a, 2, , Sol. (c) ZQAP = ZQPA = 90 - -, , ZPQB = ZPBQ = 180-6, 6, ZBCA = ZABC = ZBPC = 45 + 4, , www.jeebooks.in, or, , Hence,, , c, , aji

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www.jeebooks.in, 234, , Textbook of Trigonometry, , Now,, , n\, r, 00 \, 90° - - I + (29 - 180°) + I 45° + - I = 180°, 4, , AB = 200 m, , 0 = 55, , 7, 70 = 571, 40 = 571 - 36, tan 40 = - tan 36, 2tan26, 3tan0 - tan3’0, 1 - tan2 29, 1 -3 tan2 9, , 2, , 4, , 2t, , 1 - t2, 2t, , 2, , 3t - P, l-3t2 ’, , where t = tanO, , 1-t2,, , 4(1 - r2), , f2 -3, , (1 — t2)2 — 4t2, , 1 — 3t2, , 4(l-4t2 + 3t<) = (t2-3)(l-6t2 + t4), t‘ - 21f4 + 35t2 -7 = 0, tan20 is the root of the equation y3 - 21y2 + 35y -7=0., , • Ex. 11; An aeroplane flying horizontally 1 km above, the ground is observed at an elevation of 60° and after, 10 seconds the elevation is observed to be 30°. The uniform, speed of the aeroplane in km/h is, (a)240, (b) 240^3, , (c) 60^3, , Sol. (b) d - H cot 30° - H cot 60 0, Time taken = 10 s, , h, , • Ex. 10. The angle of elevation of towerfrom a point A, due south of it is 30° andfrom a point B due west of it is 45"., If the height of the tower be 100 m, then AB=, (a) 150 m, (b) 200 m, (c) 173.2 m, (d) 141.4 m, , Sol. (b) OB = 100 cot 45°, OA = 100 cot 30°, AB = J(OA2 + OB2), , (d) None of these, , d, , H, , 4, , 1 km, , eo°/^, < \ 30°, , _, , cot 30° - cot60°, Speed =-------------------- X 60x60 = 240^, 10, , JEE Type Solved Examples:, More than One Correct Option Type Questions, • Ex. 12 In A ABC, the ratio —-— = ■ - = —-— is, sin/4 sinB sinC, always equal to (All symbols used have usual meaning in a, triangle.), , a, , and, , (a) 2R, where R is the circumradius, where A is the area of the triangle, 2A, 2, (c) -(a2 + b2 + c2)2, 3, , (b), , (d)-^L, W2h3)', , b, ~^— = 2R ., sinB sinC, R = ^ =, 2R = ^, 4A, 2A, a, bare true, sin A, , Now,, , -(a2 + b2+c2)1/2, 3, , If B true iff a = b = c, .’. 'c is incorrect, Now, for option ‘d’, (abc)m, We have,, , (hth2h3r, , www.jeebooks.in, Sol. (a,b,d) We know that, , We know,, , -ah. = — bh2 = -ch, = A, 2*2, 2

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www.jeebooks.in, 235, , Chap 03 Properties and Solutions of Triangles, , ,, , 2A ,, , \ = — ’h2 =, , 2A ,, , zvs, , 2A, , = ------, , abc, . , ,, 8A3, h,h,h. = —, 123 abc, , (abc)213, , _ (abc)213, , (hM13, , A3, s, , (b) r, r, r. =----------------------------- = A s, ’ 2 ’ s(s-a)(s-b)(s-c), , r-, , (abc)2'3 (abc)113, , f(abc, 8A' f3 ~, , 2A, (c) Denominator =, , abc, = —= 2R, , Hence option, a, b and d are correct., , abc, Nr, , Which of the following is/are correct?, , D7” (la2}R, )R, , The value of(AC)2 is equal to 19, , =>, , The sum of all possible values of product AC • BD is, equal to 35, , (c), , The sum of all possible values of (AD)2 is equal to 29, , (d), , The value of (CD)2 can be 4, , Sol. (a,c,d) Area of quadrilateral, = 4^3 = - X2 X 3sinl20° + -xy sin60°, 2, 2, 4^ = ^, 3+—, 2, , 2, , • Ex. 14. In a&ABC, which of the following quantities, denote the area of the triangle?, , a2 - b2 sinA sinB, , 2, , (la2) abc, , abc, , R~, , = 4A, , C is not correct., , A2 s(s-a) s(s-b) s(s-c), , w7~—a—~, s(s - a) (s - b) (s - c), , A2, , A, , ” A ~, , &, , Hence, (a), (b) and (d) are correct., , • Ex. 15. Consider the system of equations, sin xcos2y = (a2 -I)2 +1 andcosx sin2y =a +1. Which of, the following ordered pairs (x, y) of real numbers can satisfy, , 16 = 6 + xy => xy = 10, D, AC2 = 22 + 32 -2-2-3 x cosl20°, x. X60\, = 4+9+6 = 19, \y, = x2 + y2 - 2xy cos 60°, A,, x2 + y2 - xy = 19, 2, or, x2 + y2 = 29, \/l20°, =>, x = 5, y = 2, 'C, 3, or, x = 2,y = 5, , (a), , R, , (a2 + b2 +c2)R, , • Ex. 13. Let ABCD be a cyclic quadrilateral such that, AB = 2,BC- 3, ZB = 120° and area of quadrilateral = 4^3., , (b), , I, , (b2 +c2 -a2)2R <, , 2bca, abc, [b2 + c2 - a2 + c2 + a2 - b2 + a2 + b2 - c2], , 2A, , (a), , A, B, Itan —tan — = 1, 2, 2, , B, 2, , A, 2, , Is2 Itan — tan — = s, , =1, , sin(A - B), , (b>^, , the given system of equations for permissible real values of, a?, , -n -n, , (a) T’ ~T, <2 2 ., , . f3n -n), , (c), , I 2, , 2 ), , (b)&Tj, W I V'T, 2 2, , Sol. (a,c,d) For permissible values of ‘a’, we must have, , (a2 -I)2 + 1 < 1 and |a + 1| < 1, => (a2 - I)2 < 0 and -1 < a + 1 < 1, =>, -2<a<0=>a:-l=0, a = 1 or - 1, , =>, , Permissible value of a = - 1, Hence, the system of equations becomes sin x cos2y = 1 and, cosxsin2y = 0., , Now, verify (a), (c), (d) alternatives., , a; +, , (c), , ?, , cot A + cotB + cotC, (..., J-2, , A, B, C, (d)r. cot—-cot—-cot—, 2, 2, 4R2 sin2 A - sin2 B^, Sol. (a,b,d) (a) —, = -sin (A + B) ab, sin(A-B) J 4R, 2, 2, , • Ex. 16. In a tsABC, let 2a2 + 4b2 + c2 = 2a(2b + c), then, , which of the following holds good?, , [Note All symbols used have usual meaning in a triangle.], (a) cos B = —, 8, , (b) sin( A - C) = 0, , www.jeebooks.in, t \ r, , = -(ab) sin C =, =A, 2, 4R, , 1, , (c)- = r, 5, , (d) sin A: sinB: sinC = 1:2:1

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www.jeebooks.in, 236, , Textbook of Trigonometry, , Sol. (b,c) We have,, 2a2 + 4b2 + c22 - 4ab - 2ac = 0, =>, (a - 2b)2 + (a - c)2 = 0, =>, a = 2b, a = c, , B, , 60°, , a =a, b, = -,c, 2, b2+c2- a2, , cos A -, , c, , a, , 2bc, 2, , a2, , 90°, , 2, , — + a,22 - a2, _4, , •c, , C ,, , 4, , J a I, 21, - a, , Hence, option (a) is not correct., a = c => A = C, , A - C = 0 => sin( A - C) = 0, , =>, , a : b: C = - :, : c = - : — : 1 = 1: ^5:2, 2 2, 2 2, Perimeter = (3 + 4i)k, (k e R), , Area of AABC = -ab, 2, , As, , 7=, 1, , -I2I 2, , A, , x., , ,, , _(s-a), , A, , y/3c, , So,, , Hence, option (b) is correct-., , (s - a)tan —, , 2, , s, , stan —, , • Ex. 18. If the length of tangents from A, B, C to the, , 2, , a, a, 4, =1--=1------ = 1-s, 5a, 5, , 1, , incircle of&ABC are 4, 6, 8, then which of the following, , 5, , is(are) correct? (All symbols used have usual meaning in a, triangle.), , 4, , (a) Area of AABC is 12^6 (b) r„ r2, r3 are in HP, , Hence, option (c) is correct., Also, a: b: c = a: -: a = 2:1:2 = sin A : sin B: sin C, 2, Hence, option (d) is not correct., • Ex. 17. In tsABC, angles A, B and C are in the ratio, , 1:2:3, then which of the following is(are) correct?, , (c) a, b, c are in AP, , 4-76, , ..., , (d) r = ——, 3, , Sol. (b, c, d) s - a = 4, s - b = 6, s - c = 8, A, A\s-a, s-a., , (All symbol used have usual meaning in a triangle)., , s-c, , (a) Circum-radius of AABC = c, (b) a : 6 : c = 1: V3 : 2, , s-b., , (c) Perimeter of AABC = 3 + 73, , B, , (d) Area of AABC =, , s-c, , s-b, , C, , s = 18, , ■J3, , A = ^18 x 4x6x8 =24^6, , Sol. (b, d) Given, A + 2A + 3A = 180°,, , B = 60° and C = 90°, R=2, , Now,, , A, , a = -,b =----2, 2, , 7, 1, Similarly, . cos B = — and cos C = —, , As, , 30°, , b, , a, , b, , c, , sin A, , sinB, , sinC, , a = 14, b = 12 and c = 10, s - a, s - b, s - c are in AP., a, b, c are in AP., A, A, A, are in HP., s-a s-b s-c, , 4^6, rz • rz are in HP r = — =, s, , 3, , www.jeebooks.in

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www.jeebooks.in, 238, , Textbook of Trigonometry, , JEE Type Solved Examples:, Statement I and II Answer Type Questions, ■ This section contains 3 questions. Each question contains, Statement I (Assertion) and Statement II (Reason)., Each question has 4 choices (a), (b), (c) and (d) out of, which only one is correct. The choices are, (a) Statement I is True, Statement II is True; Statement, II is a correct explanation for Statement I., (b) Statement I is True, Statement II is True; Statement, II is NOT a correct explanation for Statement I., (c) Statement I is True, Statement II is False., (d) Statement I is False, Statement II is True., • Ex. 22. Statement I In a AABC, ifa<b<c andr is, , inradius andrvr2, r3 are the exradii opposite to angle A, B, C, respectively, then r <r, <r2<r3., , • Ex. 23. Statement I If the sides of a triangle are 13, 14,, 15 then the radius of in circle = 4, Statement II In a &ABC, A = y/s(s - a)(s - b)(s - c) where, , a+b+c, 2, , r, s-a>s-b>s-c, s>s-a> s-b> s-c, A, A, A, A, ------- <------- <-----s-a s-b s-c, , A = V21-8-7-6 = >/3-7-24-7-3 = 3-7 -4 = 84, A 84, r=—=—=4, s, 21, 2 A, , cos, • Ex. 24. Statement I In a LABC, E — 2 has the value, a, S2, , r2, , r3, , A, , equal to----- ., abc, , 2, , r<r}<r2< r3, A, A, A, A, Statement II r, = -------> r2 =----- -> G =, -, r = —, s-a, s-b, s-c, s, , r,, , s, , A _, Statement II In a &ABC, cos —, 2, , Sol. (b) Statement I a < b < c, , Ills, , A, , Sol. (a) s = 21, , I, , Statement II For, &ABC r}r2 + r2r3 +r3r,, , ., , 5=------------- (—, and r= —, , 1_, , r, , v, , ac, , (s ~b)(s -c), , c ~ /(s-a)(s-b), ,cos —, 2 V, ab, , cos 2, 2 _s(s-a), So/, (c)---a, abc, cos2 —, Z------ 2_ s2, a, abc, , JEE Type Solved Examples:, Passage Based Questions, Passage I, , 27. Let A denote the area of the AABC and Ap be the, , (Ex. Nos. 25 to 27), , In a AABC, let tan A = 1, tan B = 2, tan C = 3 and c = 3., 25. Area of the AABC is equal to, 3 5/2, , (b)3, , (a)—, 2, (c) 2V3, , Sol, (Ex. Nos. 25 to 27) We have,, , (d) 3V2, , tan A = 1 => sin A = -7=;, , 26. The radius of the circle circumscribing the triangle, ABC, is equal to, , , . Vio, , area of its pedal triangle. If A = k &p, then k is equal, to, (a) 710, (b) 2J5, (c)5, (d) 2V1O, , tan B = 2 => sin B =, , www.jeebooks.in, (a)—, , (b)Vs, , (c) Vio, , <d)2, , 3, tanC=3=> sinC = -j=, V10

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , =>, , (a + c)2 - b2 = 2ac, a2 + c2 + 2ac - b2 = 2ac, a2 + c2 = b2, , =3, , • Ex. 37. Consider an obtuse angled triangles with side, 8 cm, 15 cm and x cm (largest side being 15 cm). If x is an, integer, then find the number of possible triangles., , 71, , So,, , Sol. (5) Since, 15 be the largest side, x2 + 82 - 152, cos0 =, 16x, for obtuse angled - 1 < cos 9 < 0, - 16x < x2 - 161 < 0, , ZB = - and ZA + ZC = 90°, 2, , Now,, sec2 A + cos2 B - cot2 C = sec2(90°-C) + cos2 90°- cot2 C, , I, , = cosec2C + 0 - cot2 C, =1, Hence, (sec2A + cos2 B - cot2C) = 1, , 15/, , Alternatively We have,, , e, , (r, +ri)-(r2 -r) = 0, , —(i), B, A, C, A- _ . _, .r, - A, B, C, As r, + r, = 4 7? sin — cos — cos — + 4/?cos — cos—sin —, ‘, J, 2, 2, 2, 2, 2, 2, B . A+C, ,B, = 47? cos —, = 4 7? cos—sin, 2, 2, 2, , -(ii), , B, C, A, A, B, C, Also, r2 - r = 47?cos— sin —cos---- 47?sin—sin— sin —, 2, 2, 2, 2, 2, 2, , A + C}, , 8, , n ■, , = 47?sin —cos, 2, , 2, , 241, , 2 B, , = 47?sin22, J, , (iii), , X, , 8, , x2 < 161, x < 13, x2 + 16x-161>0, (x + 23) (x - 7) > 0, x>7, or, 7 < x < 13, x E {8,9,10,11,12}, =>, Hence, number of possible values of x is 5., =>, and, =>, , ...(i), , •••(ii), , .‘.Using Eqs. (ii) and (iii) in Eq. (i), we get, A, , f, , J, , 42? cos---- sin — = 0, , I, , =>, , • Ex. 38. Let ABC be a right angled triangle atC. If the, inscribed circle touches the side AB to D and (AD) (BD) =11,, then find the area oftsABC., , 2J, , 2, , 47? cosB = 0, it, , ZB = - and ZA + ZC = 90°, 2, , Hence, (sec2A + cos2 B - cot2 C) = 1, • Ex. 36. In AABC, let b = 6,c = 10 and ry=r2+r3+r then, , find area of&ABC., [Note All symbols used have usual meaning in a triangle.], Sol. (30) We have,, =>, , =>, =>, , (n - r) = (r2 + r5), s - (s - a), , 2s - (b + c), , s(s - a), , (s-b) (s — c), , (s-b)(s-c), , s(s - a), , =>, Hence,, , = r2 + r3 + r, , =1, , 2A, A, o, tan2— = 1 => — = 45°, 2, 2, A = 90°, , Now, area of AABC = - be sin A, 2, , [given], , Sol. (11) We have,, (AD)(BD) = 11, =>, (s - a)(s - b) = 11, =>, (2s - 2a) (2s - 2b) = 44, =>(b + c - a) (a + c - b) = 44, c2 - (b - a)2 = 44, , b, , 2ab = 44, C, [As c2 = a2 + b2], ab = 22, 1, 1, Now, area(AABC) = -ab = -(22) = 11, 2, 2, , B, , • Ex. 39. Consider a &ABC and leta,b and c denote the, , lengths of the sides opposite to vertices A, B and C,, respecively. Suppose a = 2,b = 3,c = 4 and H be the, orthocentre. Find 15(HA)2., Sol. (196) We know that, HA = 2R cos A, where, , cos A =, , b2 + c2 -a2 _ 9 + 16-4, 2bc, ", 24, , 21, , 7, , www.jeebooks.in, = -(6) (10) sin 90°, 2, , = 30 sq units., , : --- s —, , 24, I, sin A = . 1V, , 8, 49, —, 64, , '15, , 8

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www.jeebooks.in, 242, , Textbook of Trigonometry, , Now,, , 3-72 _ 4m, , —— = 2R, sin A, , 2R^, 715, , =>, , m + 1 45/2, , 2, , 16, , 5, , 3 = -^!-, , /15, , m+1, , HA = 2R cosA = ~x-, , 715, , 3m+ 3 = 5m, 3, m=2, AB = 2m, c=3, , 8, , - _li_, V15, , Hence, 15(AH)2= 196., • Ex. 40. In a tsABC, the internal angle bisector of Z.ABC, , • Ex. 41. In &ABC has AC = 13, AB = 15 and BC =14. LetO, , meets AC at K. IfBC = 2, CK = 1 and BK = ——, then find, , be the circumcentre of the A ABC. If the length of perpendic­, ular from the point ‘O’ on BC can be expressed as a rational, , the length of side AB., , — in the lowest form, then find (m + n)., n, , Sol. (3) Using cosine law in ABKC,, , „, , D, , Sol. (41)A = Js(s - a) ($ - b) (s - c), , 4 + —-1, 4, , COS — =-------- 7=------, , 2, , = 721-8-6-7 = 784, , 2-^.2, , p_ abc _ (14) (13) (15) _ 2-13-15, 4A, 4-84, 4-12, , 2, B, , 65, , 8, A, , p = Jr2-!2, 2, , I—, ~ yL 8 I -72, 3+_____ 2 =, 6>/2, , Now,, , ------ i----- B', 15 _, , 12V2, , 5, , “7*, , ■7121-9, =--------8, 33 m, , 4V2, , 2ac, B (2-2-2m) 5, x =------ cos — - -------------a+c, 2, 2m + 2 4a/2, , ■12 ——-, , 8, , b=13 h, , 0, , _ 7(65)!-(56)!, , K, , ■4, , h-p, , c=15, , P, M, a=14, , ■+-H-, , c, , ■1 —, , n, , m + n = 33 + 8 = 41., , Subjective Type Examples, • Ex. 42. Two sides of a triangle are given by the roots of, the equation x2 - 2-^3x + 2= 0. The angle between the sides, , is, , 71, , =>, , =>, , Find the perimeter ofts.-, , Sol. We have, two sides of a triangle are given by roots of the, equation x2 - 2y/3x + 2 = 0., , =>, , 71, , a2+ b2-c 2, , 3, , 2ab, a2 + b2 - c 2, , cos — =, , 1, —=', 2, 2a b, a2 + b2 - c2 = ab ', (a + b)2 - 2ab - c2 = ab, , [using Eq. (i)j, , [using Eq. (i)], , a + b = 2V3, 71, , ab = 2 with ZC = —, 3, using cosine law, we have, a2+ b2- c2, cosC =, 2ab, , and, , (i), , 12-4-c2 =2, c2 =6; c = 76, , .'.The perimeter of A, = a + b + c = 2V3 + V6, , www.jeebooks.in

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , tan B < tan(90° - C), tan B < cotC, or, tan B • tan C < 1, If A is obtuse, then tan A tan C < L, , • Ex. 43. If in &ABC, ZA = 90° and c, sin B, cos B are, , rational numbers, then show a and b are rational., , Sol. Let AD be perpendicular from A to BC., , Then,, =>, , Similarly,, =>, , n BD, cos B = —, c, , • Ex. 46. If A is the area and 2 s the sum of the sides of a, , BD = c cos B, BD is rational., AD = csinB, AD is rational., , s2, triangle, then show A < —=., 3V3, , 2s = a + b 4- c,, , Sol. We have,, , A, , ./K, , D, AD, Now,, sinC = cosB =---b, =* b is rational., . n DC, Since,, cos C = sin B =---b, DC is rational., =>, Hence, a = BD + DC is rational., Thus, a is rational and b is rational., 8, , A2 = s(s - a)(s - b)(s - c), , Now, AM > GM, =>, , s 4- (s - g) 4- (s - fe) 4- (s - c), , >{(s-g)(s-&)(s-c)r, , ±S*£(a2)’" =>£>a,/2, 4, , Also,, , or, , (i), , 3s-2s, , >, , aT, , 3, , s, , A2^, , or, , ->, 5, , =>, , A <-^=, 3-v3, , 1/3, , or, , 5 J, , 3, , s, , 27, , ...(ii), , Thus, from Eqs. (i) and (ii);, 3<3, , [using Eq. (i)], , 2bc, , 4bc - 3b2, , • Ex. 47. In a triangle, ifr} >r2>r3, then show a >b>c., , Sol. We have,, ■=>, , =>, , s-a s-b s — c, -----<------- <------AAA, , =>, , s-a<s-b<s-c, , ., , • Ex. 45. If A, B andC are angles of a triangle such that, , n > r2 > rz, b, b, b, ------ > .------- >-------s-g s-b s-c, , 2bc, , 4c-3b, , Sol. Since, A is obtuse angle,, then, 90° < A < 180°, =>, 90° < 180 - (B 4-C) < 180°, , 3, , ^4, , 2bc, b2 +c2 - 4b2 - c2 + 4bc, , LA is obtuse, then show tan B tan C < 1., , 4, (s - g) 4- (s - fr) 4- (s - c), , >{(s-g)(s-h)(s-c)}'il'3, , Sol. Since sides of the triangle are in AP., , 2c, , 2, , A< —, , the smallest side, then express cos A in terms ofb and c., , i.e. a, b, c are in AP and let a < b < c., 2b = a + c, b2+c2- a,2, Now,, cos A =, 2bc, b2+c2- (2b - c)2, , 4, , C, , • Ex. 44. If the sides of a triangle ABC are in AP and ‘a’ is, , cos A =, , 243, , -a < - b < - c or a> b> c, • Ex. 48. ABC is a triangle and D is the middle point of, , BC. If AD is perpendicular to AC, then prove that, 2(c2-a2), , cos A ■ cos C =, , 3ac, , -90° < - (B + C) < 0, , www.jeebooks.in, 90° > B 4- C > 0, B 4- C < 90°, B<90°-C, , Sol. D is the mid-point of BC, , BD = DC = -, , 2

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www.jeebooks.in, »., , 2, , A-C, A+C, = 4 cos---------- cos--------2, 2, , „, A-C, A+C, = 2 cos---------- cos--------2, 2, , [, , [using Eq. (i)], , 2, , = 4 cos2, , 2, , b, , COS0 = ——, cos0 = ——, cos, b+c, a+c, t-n2 0 . *--2, , Sol. Given,, , COS0 = - — -, , b+c, 1-tan2_______ 2 _ a, 20, 1 + tan - b + c, 2, , of an acute angled A on the opposite sides and produced to, , meet the circumscribing circle. If these produced parts be, Ct, P, y respectively, then show that, a b c, - + - + — = 2(tan A + tan B + tan C), , f 20 b + c - a, tan — =, 2 a+b+c, , —(0, , Similarly,, , . 20_a+c-b, tan — =, 2 a+b+c, , -.(ii), , and, , a + b- c, tan — —, 2 a+b+c, , ...(iii), , a 3 y, Sol. Let AD be perpendicular from A on BC when AD is, produced, it meets the circumscribing circle at E., From question, DE =a., , , t__2 V, , tan —I-tan —I-tan —=1, 2 2, 2, , • Ex. 58. Perpendiculars are drawn from the angles A, B, C, , On adding Eqs. (i), (ii) and (iii), we get, 20, ,0, a+b+c, tan — + tan—+ tan2—=----------222 a + b + c, , Since, angles in the same segment are equal,, Z.AEB = Z.ACB = ZC, and, . ZAEC = Z.ABC = ZB, , 20, , =>, , From the right angled triangle BDE,, , tanB = —, DE, , C, , -—-—, a+b, , where 9,0 and y lie between 0 and n, prove that, , =>, , ^nC = —, DE, From the right angled triangle CDE,, , 247, , • Ex. 59. In any IsABC, if, , A+C, , _, A+C, A-C, = cos A + cosC = 2cos-------- •cos-------2, 2, A+C, 2A+C, A+C, = 2 cos, •cos-------- = 4 cos, 2, 2, 2, [using Eq. (i)], Hence Proved., , RHS, , Chap 03 Properties and Solutions of Triangles, , 20, , 2V, , tan — + tan — + tan — = 1, 2, 2, 2, , —(i), , •••(ii), , • Ex. 60. The product of the sines of the angles of a, triangle is p and the product of their cosines is q. Show that, the tangents of the angles are the roots of the equation;, , qx2 - px2 + (1 + q)x - p = 0, , Sol. From the question, sin A -sin B-cos C = p, and, cosA-cosBcosC = p, , B, , -4^, , tan A - tanB - tanC = —, 9, , Similarly,, , and, , DE, b, tanC + tanA = —, , Also, tan A + tanB + tanC = tan A-tanB-tanC, , 'C, , E, On adding Eqs. (i) and (ii), we get, BD + CD, tanB + tanC =, DE, BC a, a, , P, , c, tan A + tan B = —, Y, , —(i), , tan A + tan B + tan C = —, 9, , Now,, , (ii), , tan A tanB + tanB tanC + tanC tan A, , ..■(iii), , sinAsinB-cosC + sin B -sin C-cos A + sinC-sin A- cos B, -(iv), , cos A ■ cos B-cos C, , ...(v), , = —[sin Asin B cos C +sinC(sinB-cos A + cos B-sin A sinC)], 9, , www.jeebooks.in, On adding Eqs. (iii), (iv) and (v), we get, , — + - + — = 2(tanA + tanB + tanC), , a P Y, , = — [sin Asin B cos C + sinCsin(A + B)], 9, = — [sin A-sin B-cos C + sin2C], 9

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www.jeebooks.in, 250, , Textbook of Trigonometry, , Since A, B, C are angles of a triangle,, A, B, C, A, B, C, cot — cot — • cot — = cot —+ cot — + cot —, 2, 2, 2, 2, 2, 2, From Eqs. (ii) and (iii), , ...(iii), , Numerator (on LHS) =----- [Ecot A / 2], K, On dividing Eq. (iv) by Eq. (i), we get, (a + b + c)2, cot A/2 + cot B/2 + cot C/2, (a2 + b2 + c2), cot A + cotB + cotC, , But, , a, c, , 2, a +B, , [using Eq. (iii)], , k 2 ), , cosa /2 • cos B/2 + sina /2 • sin B/2, cosa/2- cosB/2 —sina/2-sinB/2, _ 1 + tana/2. tan B/2, 1 - tana/2. tanB/2, Write tan a /2 • tan B12 from figure;, , 1+, , 1-, , 2sin2a /2, , 1+, , 7 - cosa, 2sin2a /2, , 1-, , 1 - cosa, 7 - cosa, , 1+x, , 1 - cosa, , 1-x, , 7 - cosa, , 7 - cosa, , given x =, a, , c, , a+c, , 1+x, , 1-x, , 2, , a, , b, , c, , 1+x, , 1, , 1-x, , 1 - cosa, , 7 - cosa, , -b, , Thus, the sides are in the ratio 1 + x : 1:1 - x., • Ex. 68. In a &ABC, //tan A/2, tan B/2, tan C/2 are in, , AP. Show thatcos A,cos B, cos C are in AP., , Sol. Given tan A/2, tan B/2, tan C/2 are in AP., .*. tan A/2 - tan B/2 = tan B/2 - tan C/2, sin A/2 sin B/2, sin B/2, sinC/2, , A+C, , it, , B, , cos A/2, , 2, , 2, , 2, , sinA/2-cosB/2 - sinB/2-cos A/2, , B, B, A-C, B, = 4 sin— cos —, 2 cos —cos, 2, 2, 2, , cos B/2, , cos B/2, , cosC/2, , cos A/2-cos B/2, sin B/2 • cos C/2 - sin C/2 • cos B/2, , cosB/2-cosC/2, , [using Eq. (i)], , A-B, B-C, sin, sin, 2 J— = --- <__ 2_, —V-------A, C, cos —, cos—, 2, 2, A, B+C, cos — = sin, 2, 2, C, • I A+B, cos — = sinl, 2, ', 2, , a, cos —, 2, sin—=, 2, , but, , and, , cos a/2, , sin, , 5/4 - cos2a /2 = 14 - 1 + cosa, 2, , cos, , a-B, , On dividing numerator and denominator by cosa/2.cosB/2, , Sol. Let AABC be the given A in which A is the greatest and, C is the least angle., Then, according to the hypothesis, ...(i), A - C =a, But, A +C = it - B, •••(ii), .".From Eq. (i) and Eq. (ii), we get, it a-B, A =-+, 2, 2, a+B, ...(iii), and, 2, 2, Again by hypothesis,, a + c-2b, .*., sin A + sin C = 2sin B, A-C, . B, B, A +C, = 4 sin— cos —, => 2sin, cos, 2, 2, 2, 2, , 2sin —, 2, , sin A, sinC, , -(iv), , • Ex. 67. If the sides of a triangle are in AP, and its great­, est angle exceeds the least angle by Ct, show that the sides, 1 -cosa, are in the ratio 1 + x: 1:1 - x, where x =, 7 - cos a, , =>, , cos, , = sin, , A-B, , B-C, , 2, ■sin, , •sin, , -0), , -(ii), , A+B, 2, , B+C, , [from Eqs. (i) and (ii)], , www.jeebooks.in, 7 - cosa, , 2, , 2, , 2, , cos B - cos A = cos C - cos B, Hence, cos A, cos B, cos C are in AP.

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 255, , i, , „2cos —cos ( -B-C, 2 V 2, , I, , C-A, 2cos —cos 2, sin A, sinB, C, A-B, 2sin —cos, 2, 2, +-----------sinC, B+C, B-C, A+C, C-A, 2sin, cos, 2sin, cos, 2, 2_, 2, 2, +----sin A, sinB, A + B^|, A-B, cos, 2sin, 2, 2, +----sinC, sin B + sin C i, sin C + sin A, sin A + sin B, +, +, sinA, J, sinB, sinC, sinB 4--------sin A >, -------, , sin A, , as, =>, =>, , sin B, , 8xyz <abc, , +—2—L, , pinC +-----sinB 'I fsin A, sinC ) LsinC, , sinC, +-----sin A, , A, B, C are angles of triangle, 0 < A, B, C < n, sin A, sin B, sin C > 0, E>2+2+2, , (s - a)(s - h)(s - c) < -(abc), 8, , s(s - a)(s - b)(s - c) < -s(abc) < —(a + b + c)(abc), 8 ., 16, , A < — fabc(a + b + c) and equality holds if, 4, x = y = z=>a = b = c, Aliter RHS = —yjabc(a + b + c), 4, , = ij2s-4BA = -J8.--RA, 4, 4V r, , ‘, , ■‘S’, , 1, 2, , as x + — > 2, if x > 0, x, , cos, , A-B, 2, , - cos, , 2f A-B, 1 1, -cos, 2 4, , I 2, , E>6, , /1 + B, , Since, sum of two sides is always greater than third side., b + c - a, c + a - b,, , a + b-c>0., =>, ($ - a)(s - b)(s - c) >0, let (s - a) = x, (s - b) = y, (s - c) = z, Now,, x + y = 2s — a - b = c, y + z = a, and, z+x=b, Since, AM > GM, then 2-Jxy < x + y = c;, 2^yz <y + z = a-,2-fzx <z + x-b, , A+B, 2, , 1, | A-B, X —cos, 2, 1, 2, , < - cos2, 8, , show that A < - y/(a + b + c)abc. Also, show that equality, 4, occurs in the above inequality if and only ifa=b = c., , Is •, = d-(b + c - a)(c + a - b)(a + b- c), V8, , •cos, , 2, , • Ex. 81. If A is the area of a A with side lengths a, b, c then, , Sol. We know, A = fs(s - a)(s - b)(s - c), , (i), , I . A . B . C, 8sm—.sin—.sin —, 2, 2, 2, .,, . A . B . C, Consider, sm — • sin — • sin —, 2, 2, 2, , A-B, , 2, , si, 8, , ,(ii), , o . A . B . C, 8sin—.sin—, sin—, 2, 2, 2, , 1, Thus, from Eqs. (i) and (ii); A < —, , + b + c)abc, , 4 ’, , A-B, 1, and equality holds, x = - cos, 2, 2, and, , cos, , A-B, 2, , =1, , A = B and cos, , A+B, , 1, , 2, , 2, , C 1, sin— = - => C = 60°, 2 2, A = B = 60°, Thus, the equality holds of triangle is equilateral, , www.jeebooks.in

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www.jeebooks.in, g Properties and Solutions of Triangles Exercise 1:, Single Option Correct Type Questions, , 1. In the adjoining figure, the circle meets the sides of an, equilateral triangle at six points., , 8. Two medians drawn from the acute angles of a right, 7t, angled triangle intersect at an angle —. If the length of, 6, the hypotenuse of the triangle is 3 units, then the area of, the triangle (in sq units) is 4~K, then K is, , (B)^, , (a) 3, , C, , E, , D, , If AG = 2, GF = 13, FC = 1 and HJ = 7, then DE equals to, (a) 2-^22, (b)7>/3, (c) 9, (d) 10, 2. In a AABC, if a = 13, b = 14 and c = 15, then ZA is equal, to (All symbols used have their usual meaning in a, triangle.), .3, , .4, , (a)sin“1-, , (b)sin~, (b) sin' -, , ,3, , (c)sin, (c) sin-1-, , (d)sin, (d) sin, , ,2, , 3. In a AABC, if b = (45 -1) a and ZC = 30°, then the value, of (A - B) is equal to (All symbols used have usual, meaning in the triangle.), (a) 30°, (b) 45°, (c) 60°, (d) 75°, , 4. In a AABC, if Z.C-105°, ZB = 45° and length of side, AC - 2 units, then the length of the side AB is equal to, , (a) 45, , (b) 45, , (c) 45 +1, , (d) 45 +1, , 5. If P is a point on the altitude AD of the AABC such that, D, , a, , b, , 10. Let A = P, 1, , ?, 1, , 45, (b)22., 8, , c, , r and B = A 2, 1, , If(a-b)2 + (p-q)2 =25,(b-c)2 +(?-r)2 =36 and, , (c - a)2 + (r - p)2 = 49, then det B is, (a) 192, (b) 864, (c) 3456, (d) 25 X 36 X 47, 11. If in a AABC, the incircle passing through the point of, intersection of perpendicular bisector of sides BC, AB,, ' A . A . B . C, ., then 4 sin — sm — sm — equals to, 2, 2, 2, , (a) 45, , (b) 45-1, , (c)V2 + l, , (d)l, 2, , 12. If two sides of a triangle are roots of the equation, x2 - 7x + 8 = 0 and the angle between these sides is 6(f,, , (d)8, , [Note All symbols used have usual meaning in AABC.], 5, , 9. If in a right angle AABC, 4 sin A cos B -1 = 0 and tan A, is finite, then, (a) angles are in AP, (b) angles are in GP, (c) angles are in HP, (d) None of these, , A, (b) 2b sin—, 3, C, (d) 2c sin—, 3, , 6. In AABC, if2b = a + c and A - C = 90°, then sin B equals, , 45, (a)2L£, , (d) None of these, , then the product of in-radius and circum-radius of the, triangle is, , ZCBP = —, then AP is equal to, , C, (a) 2a sin —, 3, B, (c) 2c sin —, 3, , (c)V3, (c)J3, , 45, (c)l£, , 45, (d)^-, , 4, , 3, , 7. Let ABC be a right triangle with length of side AB = 3, , and hyotenuse AC = 5. If D is a point on BC such that, BD, AB ., ., ---- =----- , then AD is equal to, DC AC, , 71, , 13. If median AD of a triangle ABC makes angle — with side, 6, BC, then the value of (cot B - cot C)2 is equal to, (a) 6, (c) 12, , (b)9, (d) 15, , 14. If the perimeter of the triangle formed by foot of, altitudes of the AABC is equal to four times the, circumradius of AABC, then AABC is, (b) equilateral triangle, (a) isosceles triangle, (d) None of these, (c) right angled triangle, , www.jeebooks.in, (b) ~~, 0»^, , (c)^, ef, , (d)^, 4

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 15. In a triangle with one angle —, the lengths of the sides, , form an A.P. If the length of the greatest side is 7 cm, the, radius of the circumcircle of the triangle is, . . 7^, _ . 573, . . 273, ... /(a)---- cm (b)----- cm (c)----- cm (d) V3 cm, 3, , 3, , 3, , 16. Sides of a triangle ABC are in AP. If a < min {b, c}, then, cos A may be equal to, 3c - 4b, ... 3c - 4b, (b)---(a), 2b, 2c, 4c -3b, 4c - 3b, (c) ~2b~, (d) —, (i), 17. The product of the sines of the angles of a triangle is p, , and the product of their cosines is q. Then, the tangents, of the angles are the roots of the equation, (a) qx3 - px2 + (1 + q)x- p = Q, (b) qx3 - px2 - (1 - q)x - p = 0, (c) qx3 - px2 + (1 + q)x + p = 0, (d) None of the above, 18. Let C be incircle of AABC. If the tangents of lengths, t1( t2 and t3 are drawn inside the given triangle parallel, , to sides a, b and c, respectively, then — + — + — is, a, b, c, equal to, (a)0, (b)l, (c) 2, (d) 3, 19. If the sine of the angles of AABC satisfy the equation, , c3x3 -c2(a + b + c)x2 +lx + m = Q, (where a, b, c are the sides of AABC), then AABC is, (a) always right angled for any I, m, (b) right angled only when, / = c(ab + be + ca) = c "Lab, m = -abc, (c) right angled only when, . cLab, abc, I =----- , m =-----4, 8, (d) never right angled, 20. In a triangle ABC, medians AD and CE are drawn. If, 71, , 77, , AD = 5, Z.DAC = — and Z.ACE = —, the area of the, 8, 4, triangle is, /(a)' ?, 50, na, 25, /(c)x 25, (b)?, , w7, , 21. In a triangle ABC, a>b>c.If, , a„3 +, b.3 +, c 3, , sin3 A +sin3 B + sin3 C, , = 8, then the maximum value of a, , 257, , 22. A triangle ABC exists such that, (a) (b + c + a)(b + c - a) = 5bc, (b) the sides are of lengths 719. -J38, 7116, , (c), , (b2-c2>|, \, , (d) co:, , «2 J, , +, , (B-C, , k 2, , c2 - a, b2, , +, , fa2-b2> = 0, ., , ,, , c2, , B + Cl, , = (sinB + sinC) co:, , 2 J, , 23. If a AABC, a, b, A are given and blt b2 are two values of, the third sides b such that b2 = 2b}. Then, sin A is equal to, !9a2-c2, 9a2-c2, (a)^, (b), 8a2, 8c2, ^9a2 - c2, (d) None of these, (c), 8b2, i, , 24. In a triangle ABC, if cot A = (x3 + x2 + x)2,, i, , cot B = (x + x +1)2 and cotC = (x, then the triangle is, (a) equilateral, (b) isosceles, (c) right angled, (d) obtuse angled, 25. In a AABC, a, b, A are given and cp c2 are two values of, , the third side c. The sum of the areas two triangles with, sides a, b, c1 and a, b, c2 is, , (a) -b2sin2A, 2, , (b) -azsin2A, 2, (c) b2sin2A •, (d) None of these, 26. In AABC, if a = 10 and b cot B + c cot C = 2(r + R\ then, the maximum area of AABC will be, (a) 50, (b) 750, (c) 25, (d) 5, , 27. Three circles touch one-another externally. The tangents, at their point of contact meet at a point whose distance, from a point contact is 4. Then, the ratio of the product, of the radii to the sum of the radii of circles is, (a) 16 : 1, (b) 1:16, (c) 8 :1, (d) None of these, , 28. Let a, b, c be the sides of a triangle. No two of them are, equal and X G R. If the roots of the equation, x2 + 2(a + b + c)x + 3X(ab + be + ca) = 0 are real distinct,, then, , (a)X<l, , (b)X>|, , www.jeebooks.in, is, , <4, , (b)2, , (c)8, , (d) 64, , . , , fl 5', (c)Xe, , <3 3., , k3 3.

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www.jeebooks.in, 258, , Textbook of Trigonometry, , !, , 29. In triangle ABC, if P, Q, R divides sides BC, AC and AB,, , respectively, in the ratio Ac: 1 (in order). If the ratio, < area APQR', is -, then k is equal to, ^area AABC,, , (b)2, (d) None of these, , (c)3, , 30. Let f(x + y) = f(x)-f(y)for all x and y and /(I) = 2. If in, , 36. If in a triangle 1 - —, r2, , ., , (a) right angled, (c) equilateral, , 1 - — = 2, then the triangle is, ri), , (b) isosceles, (d) None of these, , 37. If the median AD of a triangle ABC makes an angle 0, with side AB, then sin(A - 0) is equal to, b, c, (a)—sin0, (b) — sin0, c, , b, , Q, , a triangle ABC, a = f(3), b = /(l) + /(3), c = /(2) + /(3),, (a)C, (b)2C, (c)3C, (d)4C, , 31. Let a, b, c be given positive numbers, then values of x, y, , 38. In a AABC, angles A, B, C are in AP, then, , ^3 - 4 sin A sin C, lim, , and z E R+ which satisfies equations x+y+z=a+b+c, , A—>C, , and 4xyz = ~(a2x + b2y + c2z) = abc are respectively., , (a) 1, (c) 3, , abc, (b) 2’ 2’2, , , ,b+c a+c a+b, (a) "T-•, • “T~, 2, 2, 2, a+b a+c b+c, , (c), , 32. If 7/, *t2’ and ‘t3’ are the lengths of the tangents drawn, from centre of x-circle to the circumcircle of the AABC,, 1, then — + — + — is equal to, ', , 'l, , t2, *3, , <2, , abc, (a)---------a+b+c, a+b+c, (c), 2abc, , a+b+c, abc, 2a be, (d), a+b+c, , (b), , 33. In triangle ABC, Z-A > —.AAt and AA2 are the median, 2, and altitude, espectively. If ZBAAj = ZAtAA 2, = ZA 2 AC, then sin 3 — • cos y is equal to, , 4b2c, , (b) X > 6, (d) X >4, , 40. In the triangle ABC, if (a2 + b2)sin(A - B), = (a2 - b2 )sin(A + B), then the triangle is, , (a) either isosceles or right angled, (b) only right angled, (c) only isosceles triangle, (d) None of the above, , 2, 2, 18°, ------+------ +------ =------- , then the maximum value of, 1!9! 3!7! 515! (2b)!, tan A tan B is equal to, , (a);, 4, , (b)l, , (c)l, (c)~, , (d)l, , 4, , 5, , 0, , equation a(b - c)x2 + b(c - a)x + c(a - b) = 0 are equal,, , then sin2, , A, , 2, , 34. In an ambiguous case of solving a triangle when a = 5/5,, , It, b = 2, ZA = — and the two possible values of third side, arecj andc2, then, (a)|Cj-c2| =2^6, , (bllq-c2| = 4a/6, , (c)|c1-c2| = 4, , (d)|q-c2|=6, , 35. If, , (a) X < 0, (c) 0 < X < 4, , 42. If a, b, c be the sides of a triangle ABC and if roots of the, , (b) 3q3, (b), 64b2c, (d)-^, (d), 12b2c, , —, 16b2c, (c)^j-, , (b)2, (d)4, , 41. In a AABC, sides a, b, c are in AP and, , 71, , (a), , is, , 39. In a triangle ABC, (a + b+ c)(b + c - a) = Xbc if, , (d) None of these, , 2’2’2, , (d) None of these, , (c)-cos0, b, , is the circumradius of the pedal triangle of a given, , triangle and R2 is the circumradius of the pedal triangle, of the pedal triangle formed, and so on R3, R4..., then, , (a) AP, (c) HP, , sin2, , B, , sin2, , 2, , 2J, , are in, , (b) GP, (d) AGP, , 43. The ratio of the area of a regular polygon of n sides, inscribed in a circle to that of the polygon of same, number of sides circumscribing the same circle is 3 :4., Then, the value of n is, (a) 6, (b)4, (c) 8, (d)12, , j-^sin22 A +sin A + 1, 44. In any AABC,, , is always greater, , sin A, , www.jeebooks.in, Rt, where R (circumradius) of AABC is 5, , the value of, , t =1, , is, (a) 8, , (b) 10, , (c) 12, , (d) 15, , than, (a) 9, (c) 27, , 7, , (b)3, (d) None of these

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 45. If the incircle of the triangle ABC, passes through it’s, circumcentre, then the cos A + cos B + cos C is, (a)-2, (b)72, (c)-Ji, (d) None of these, , 46. The perimeter of a triangle is 6 times the arithmetic, mean of the sines of its angles. If the side a is 1, then A is, equal to, (a) 30’, (b) 60’, (c) 90’, (d) 120’, 41. If there are only two linear functions f and g which map, , [1, 2] on [4, 6] and in a AABC, c = /(I) + g(l) and a is the, maximum value of r2, where r is the distance of a, variable point on the curve x2 +y2 - xy = 10 from the, origin, then sin A: sin C is, (a) 1 : 2, (b) 2 : 1, (c) 1 :1, (d) None of these, , 48. A circle is inscribed in an equilateral triangle of side a., The area of any square inscribed in this circle is, / \, , (a) a, , 2, , \ &, , 2, , (b) —, 4, , z x fl, , 2, , c)—, 3, , /i\fl, , 2, , (d) —, 6, , 49. In any triangle ABC, if sin A, sin B, sin C are in AP, then, , the maximum value of tan — is, 2, , (d) None of these, sin B, ,„ 2, 50. In a AABC, 2 cos A = ------ and 2 tan B is a solution of, sinC, equation x2 - 9x + 8 = 0, then AABC is, , (a) equilateral, (c) scalene, , (b) isosceles, (d) right angled, , 51. A triangle is inscribed in a circle. The vertices of the, triangle divide the circle into three arcs of length 3, 4, and 5 unit, then the area of the triangle is equal to, , . 973(1 + 73), 973(73 -1), (a)------- ------ sq unit, (b)------- ------ sq unit, n, n, 973(1 + 73), 973(73 -1), sq unit, sq unit, (c), (d), 2n2, 2n2, , 259, , 31, 54. If in a triangle ABC, a = 5, b = 4 and cos(A - B) = —,, 32, then the third side c is equal to, (a) 3, (b) 6, (c) 7, (d) 9, , 55. In AABC, if AB = x, BC = x + 1, Z.C = —, then the least, 3, integer value of x is, (a) 6, (b)7, (c) 8, (d) None of these, 56. Three circular coins each of radii 1 cm are kept in an, equilateral triangle so that all the three coins touch each, other and also the sides of the triangle. Area of the, triangle is, , (a) (4 + 273) cm,2!, , (b) — (12 + 773) cm.2:, 4, , (c) —(48 + 773) cm',2, 4, , (d) (6 + 473) cm.22, , 57. The sides of a triangle are in AP. If the angles A and C, are the greatest and smallest angle respectively, then, 4(1 - cos A)(l - cos C) is equal to, (a)cosA-cosC, (b)cosAcosC, (c)cosA + cosC, (d)cosC-cosA, ', ., 58. If in AABC, c(a + b) cos - B = b(a + c) cos - C, the triangle, 2, 2, is, (a) isosceles, (b) equilateral, (c) right angled but not isosceles, (d) right angled and isosceles, 59. In a triangle, the line joining the circumcentre to the, incentre is parallel to BC, then cos B + cos C is equal to, (a)^, , (b)l, , 3, (0)7, 4, , 1, (d)2, , 60. In the given figure, AB is the diameter of the circle,, centered at 0. If Z.COA = 60°, AB = 2r, AC = d and, CD = /, then I is equal to, fl, ■D, , 52. If a, b and c are the sides of a triangle such that b-c = A2,, , then the relation in a, X and A is, (C, (a) c > 2Xsin —, (b) b > 2Xsin, I2, (c) a > 2Xsin, , 0, , C, , (d) None of these, , A, , 53. In AABC, AD is an altitude from A on side BC. lfb>c,, , www.jeebooks.in, Zc = 23° and AD = -^~, — .then ZB is, b2— c 2, , (a) 110’, , (b) 113*, , (c) 120’, , (d) 130’, , (a)d73, (c)3d

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www.jeebooks.in, 260, , Textbook of Trigonometry, , !, !, , the foot of the tower. If a pole mounted on the tower, also subtends an equal angle at O, the height of the pole, is, , 61. In a triangle ABC; AD, BE and CF are the altitudes and R, is the circum radius, then the radius of the circle DEF is, (a) 2R, (b) R, D, , (c) —, 2, , (2, , (d) None of these, , 62. In a right angled triangle ABC, the bisector of the right, angle C divides AB into segment x and y and, A-B, = t, then x: y is equal to, tan, 2, , (a)(l + t)=(l-t), (c)l:(l + t), , (b)(l-t):(l + t), (d)(l-t):l, , •, , 63. A variable triangle ABC is circumscribed about a fixed, circle of unit radius. Side BC always touches the circle at, D and has fixed direction. If B and C vary in such a way, that (BD)-(CD) = 2, then locus of vertex A will be a, straight line, (a) parallel to side BC, (b) right angle to side BC, , (c) making and angle — with BC, 6, , (d) making an angle sin, , (a)b, , ( a2 + b2], , (b)&, , y-b2;, ra2 + b2), (d)a, , y^a + b, („2, , (c)a, , .2, , a —b, , „2 , >2, 12, , a —b, , ly-*2,, , y^a 2 +, b12, , i, , 66. A balloon is observed simultaneously from three points, A, B and C on a straight road directly under it. The, angular elevation at B is twice and at C is thrice that of, A. If the distance between A and B is 200 m and the, distance between B and C is 100 m, then the height of, balloon is given by, (a) 50 m, (b) 50^3 m, (c) 50^2 m, (d) None of these, 67. A vertical pole (more than 100 m high) consists of two, portions, the lower being one third of the whole. If the, 1, at a point in, upper portion subtends an angle tan, 2, , a horizontal plane through the foot of the pole and, distance 40 ft from it, then the height of the pole is, (a) 100 ft, (b) 120 ft, (c) 150 ft, (d) None of these, , - | with BC, ,3j, , 65. A tower of height b subtends an angle at a point 0 on, the level of the foot of the tower and at a distance a from, , § Properties and Solutions of Triangles Exercise 2:, More than One Correct Type Questions, 68. If area of AABC, A and ZC, are given and if the side c, opposite to given angles is minimum, then, I 2A, I 2A, (a)a =, (b)h =, sinC, sinC, 4A, sinC, , sin C, , 69. If A represents area of acute angled AABC, then, yja2b2 -4A2 + 7^2c2 - 4A2 + -Jc2a2 - 4A2 equals to, , (a) a2 + b2 + c2, (b), , a2 + b2 + c2, 2, , (c) ab cosC + be cos A + ca cosB, , (d) ab sinC + be sin A + ca sinB, , 70. In AABC, the value of c cos(A - 0) + a cos(C + 0) cannot, , exceed (0 6 (0,2k)) [Letters have usual meanings], (a) a, (b) b, (c) c, (d) s., , 7t, , (a) measure of ZA is —, 2, 71, , (b) measure of ZB is —, 2, C, r-Jl, (c) cot — = V7, 2, , 2, (d) circumradius of AABC is —, —, yl/4, , 31, 72. In AABC, let a = 5, b = 4 and cos( A - B) = —, then which, 32, of the following statement(s) is (are) correct?, [Note All symbols used have usual meaning in a, triangle], , (a) The perimeter of AABC equals —, 2, V7, (b) The radius of circle inscribed in AABC equals —, 2, , www.jeebooks.in, 71. In AABC, if ac = 3, be = 4 and cos( A - B) = -, then, 4, , (c) The measure of ZC equals cos-1 8, (d) The value of R(b2 sin2C + c2 sin2B) equals 120

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 73. In which of the following situations, it is possible to, have a AABC?, , (All symbols used have usual meaning in a triangle), (a) (a + c - b) (a - c + b) = 4bc, (b) b2sin2C + cos2sin2B = ab, , M+i, 2, , V3, cos A cos B = — = sin Asin B, 4, , 79. Let a, b, c be the sides of triangle whose perimeter is P, , 74. In a AABC, let BC = 1, AC = 2 and measure of ZC is 30°., i, , 78. There exist a triangle ABC satisfying, (a) tanA + tanB + tanC = 0, _. sin A sinB sinC, (b) “~, =, 2, 3, 7, (c) (a + b)2 = c2 + ab and <2 (sin A + cos A) = ^3, , (d)sinA + sinB =, , (c) a = 3, b = 5, c = 7 and C = —, 3, A-C, A + C^, (d) cos, = cos, 2, 2, , 261, , Which of the following statement(s) is(are) correct?, (a) 2sinA =sinB, (b) Length of side AB equals 5 - 2yf3, , (c) measure of ZA is less than 30’, (d) Circumradius of ZABC is equal to length of side AB, , 75. Let one angle of a triangle be 60’, the area of triangle is, 10>/3 and perimeter is 20 cm. If a > b > c where a, b and c, , denote lengths of sides opposite to vertices A, B and C, respectively, then which of the following is(are) correct?, (a) Inradius of triangle is 5/3, (b) Length of longest side of triangle is 7, 7, (c) Circum-radius of triangles is, , and area is A, then, (a) P3 < 27(b + c - a)(c + a - b)(a + b - c), (b) P2 <\a2 + b2 + c2), , (c) a2 + b2 + c2 > 4^3A, (d) P* < 25 < A, , 80. If in AABC, A = 90° and c, sin B and cos B are rational, , number, then, (a) a is rational, (b) a is irrational, (c) b is rational, (d) b is irrational, 81. In AABC, which of the following statements are true, (a) maximum value of sin2A + sin2B + sin2C is same as the, maximum value of sin A + sinB + sinC, (b) R > 2r, where R is circumradius and r is inradius, , (d) Radius of largest escribed circle is —, 12, , (a + b + c), , 76. In a AABC, if a = 4, b - 8 and ZC = 60°, then which of the, , following relations is(are) correct?, , [Note All symbols used have usual meaning in triangles, ABC], (a) The area of AABC is 85/3, , (d) AABC is right angled if r + 2R = s, where s is, semiperimeter, , 82. If I is the length of median from the the vertex A to the, side BC of a AABC, then, (a) 4/2 =2bz + 2c2-a2, , (b) The value of Esin2 A = 2, , (b) 4/2 = b2 + c2 + 2bccosA, , 2^3, (c) Inradius of triangle ABC is, 3 + V3, , (c) 4l2 = a2 + 4bccosA, , •4, (d) The length of internal angle bisector of ZC is -7=., , (d) 4l2 -(2s - a)2 - 4bcsin2^^, , V3, , 77. Given an isosceles triangle with equal sides of length b,, 71, , 83. If A, Ap A2, A3 are the areas of the inscribed and, escribed of a AABC, then, , base angle a < — and R, r the radii and 0,1 the centres of, , (a) Ta + Ta + Ta=+r2 + r3), , the circumcircle and incircle, respectively. Then, , 1, 1, 11, (b) -r^ + -n= + -rr = -7T, , (a) R = ~b cosec a, 2, (b) A = 2b2sin2a, fesin2a, (c)r =, 2(1 + cos a), , . . i, 1, 1, s2, (c) -= + -= + -= = -=----, , TA TA TA ■Jnrlrfi, (d) Ta + Ta + Ta = Vit(4K+o, , 84. If a, b, A be given in a triangle and q and c2 two possible, PCOS ----, , V2J, „2smacos, ., f —a A, , values of third side such that cf + qc2 + c2 = a2, then, , www.jeebooks.in, (d)OZ =, , <2 ), , A is equal to, (a) 30*, (c) 90*, , (b) 60*, (d) 120*

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www.jeebooks.in, 262, , Textbook of Trigonometry, , 85. D, E and F are the middle points of the sides of the, , triangle ABC, then, (a) centroid of the triangle DEF is the same as that of ABC, (b) orthocentre of the triangle DEF is the circumcentre of ABC, (c) orthocentre of the triangle DEF is the incentre of ABC, (d) centroid of the triangle DEF is not the same as that of ABC, , 86. The sides of AABCsatisfy the equation2a2 + 4fe2 + c2, , - lab + 2ac. Then, (a) the triangle is isosceles, (b) the triangle is obtuse, -if 7, (d) A = cos', (c) B = cos k8, 87. If A represents the area of acute angled triangle ABC,, then ^a2b2 - 4A2 + *jb2c2 - 4A2 + ^c2a2 - 4A2 is, , 89. If H is the orthocentre of triangle ABC, R = circumradius, , and P = AH + BH + CH, then, (a) P=2(R + r), (b) max. of P is 3R, (c) min. of P is 37?, (d) P = 2(R - r), , 90. If inside a big circle exactly n(n > 3) small circles, each, of radius r, can be drawn in such a way that each, small circle touches the big circle and also touches, both its adjacent small circles, then the radius of big, circle is, 71, , 1 + tan—, n, (b), n, cos—, \, n /, ' . n, 2ti jz, r sm— + cos—, 2n, (d), . n, sin—, n, , , . f, n, (a) r 1 + cosec —, \, n., , equal to, , a 2 +, t)h2 +, c 2, 2, (c) afecosC + be cos A + ca cos B, (d) afesinC + fecsinA + casinB, , (a) a2 + b2 + c.2‘, , (b), , 88. In triangle, ABC, if2a2b2 + 2b2c 2 = a4 +fe4 + c4,then, angle B is equal to, (b) 135’, (d) 60’, , (a) 45’, (c) 120’, , / x, , 271, , (c) r 1 + cosec —, n, , 91. If in triangle ABC, a, b, c and angle A are given and, csin A < a < c, then (b} and fe2 are values of fe), (a) fe, + b2 = 2ccosA (b) fej + b2 = ccosA, (c) fejhz = c2 - a2, (d) blb2 = c2 + a2, , g Properties and Solutions of Triangles Exercise 3:, “ Statement I and II Answer Type Questions, ■ This section contains 15 questions. Each question, contains Statement I (Assertion) and Statement II, (Reason). Each question has 4 choices (a), (b) and (d), out of which only one is correct. Choices are, (a) Both Statement I and Statement II are correct and, Statement II is the correct explanation of Statement I, (b) Both Statement I and Statement II are correct and, Statement II is not the correct explanation of, Statement I, (c) Statement I is correct but Statement II is incorrect, (d) Statement II is correct but Statement I is incorrect, 92. In a AABC,, a3 + b3, , jr, , Statement I ZB = —, 2, , 2, Statement II sin A = —r=, , V5, , 94. Statement I If in a triangle ABC sin2 A + sin2, , B + sin2 C = 2, then one of the angles must be 90’., Statement II In any triangles ABC, , cos 2A + cos 2B + cos 2C = -1 - 4 cos A cos b cos C, 95. Statement I If A, B, C, D are angles of a cyclic, , quadrilateral then Esin A = 0., +c33, , 2, , =c2(a + b + c), , (All symbol used have usual meaning in a triangle.), Statement I The value of ZC = 60°., Statement II AABC must be equilateral., , Statement II If A, B, C, D are angles of cyclic, quadrilateral then, Seos A = 0., 96. Statement I In any triangle ABC, the square of the, , length of the bisector AD is fee 1------ - ----- ., , I, , 93. In a AABC, let a = 6, b = 3 and cos(A - B) =, , (i> + c)V, , www.jeebooks.in, [Note All symbols used have usual meaning in a triangle.), , Statement II In any triangle ABC length of bisector AD, 2bc, | A, is-------- cos, , (fe + c), , 2

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www.jeebooks.in, 263, , Chap 03 Properties and Solutions of Triangles, , 97. Statement I If I is incentre of AABC and excentre, opposite to A and P is intersection of and BC, then, IP-I}P= BP-PC, , Statement II In a AABC, I is incentre and is, excentre opposite to A, then £BI, Ip C must be square., , i, , 98. All the notations used in statement I and statement II are, usual., pnr /I, cos B cosC, Statement I In triangle ABC, if---- —, ~b~, c, a, r, i+, r, 2, +, r, 3, then value of, is equal to 9., r, b, —-2R,, Statement II If AABC : ——, sin A sinB sinC, where R is circumradius., 99. Statement I In a triangle ABC if tan A : tan B: tan C = 1, : 2 : 3, then A = 45°, , Statement II If p : q: r = 1:2:3, then p = 1, 100. Statement I In any right angled triangle, , a2 + b2 + c 2, — is, R2, , always equal to 8., , Statement II a2 = b2 + c2, 101. Statement I perimeter of a regular pentagon inscribed, in a circle with centre 0 and radius a cm equals, ,10asin36° cm., , Statement II Perimeter of a regular polygon inscribed, in a circle with centre 0 and radius a cm equals, (360°, (3n - 5) sin ----- cm, then it is n sided, where n > 3., I 2n ), , 102. Statement I In any triangle ABC, a cos A + b cos B + c cos C <s., , Statement II In any triangle ABC, B . fCV, . I A ] . ( B}, C, 1, sinl — |sin| — | sm| — | < 8, V2, 2, 2, , ABC, 103. Statement I In a AABC, if cos2 — + cos2 — + cos2 —, 2, 2, 2, (2, , =, , \, , 1, , ,, , •, , c, , 1, , ■, , 9, , x + — L then the maximum value or y is -., , 8, , x2 J, , Statement II In a AABC,sin — -sin— sin— < 2, 2, 2 8, 104. Statement I In any triangle, a cos A + b cos B + c cos C < s, A) . (B, B, Statement II In any triangle sin — |sin| —, 2, , C, 2, , <1, 8, , q2, + b2 + c, 105. Statement I In triangle ABC, — __, , Statement II If at > 0, i = 1,2,3,..., n which are not, identical, then, a\ + a2 +..,+ an > ai +a2 +...+an, n, n, m>L, , m, , , if m < 0 or, , 106. Statement I AAP BBP C^ are the medians of triangle, ABC whose centroid is G. If the points A, Cp G and B are, concyclic, then c2,a2,b2 are in AP., , Statement II BG • CCX = BCj • BA, , H Properties and Solutions of Triangles Exercise 4:, Passage Based Questions, Passage I, (Q.Nos. 107 to 109), , R is circumradii of AABC, H is orthocentre. Rl,R2, B3 are, circumradii of AAHB, AAHC, ABHC. If AH produced meet, the circumradii of ABC at M and intersect BC at L, ZAHB = W)° -C, c, = 2Ri, sin(180°-C), , 107. R}R2 + R2R3 + Rx R3 is equal to, (a)2R2, (b)3J?2, (c)5R2, (d) R2, , 108. Area of AAHB, (a) 2R cosA cosB cosC, (b) R2 cos A cosB cosC, , (c) 2R2 cos A cosB sinC, (d) None of the above, , www.jeebooks.in, c, = 2Rl, sinC, 2?! =R, , 109. Ratio of area of AAHB to ABML, is, (a) cosB : 2 cos A, (b)2:l, (d) None of these, (c)cosA:cosB cosC

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www.jeebooks.in, 264, , Textbook of Trigonometry, , Passage II, , Passage IV, , (Q. Nos. 110 to 112), , (Q.Nos. 116 to 118), , Let ABC be an acute triangle with BC = a,CA = b and AB = c,, where atbtc. From any point 'p' inside &ABC let D, E, F, denote foot of perpendiculars from ‘P’ onto the sides BC,CA, and AB, respectively. Now, answer the following questions., HO. All positions of point ‘P for which ADEF is isosceles lie, on, (a) the incircle of 6ABC, (b) line of internal angle bisectors from A, B and C, (c) arcs of 3 circles, (d) None of the above, 111. Let A(7,0), B(4,4) and C(0,0) and ADEF is isosceles with, DE = DF. Then, the curve on which ‘F may lie, (a) x = 4 or x + y = 7 or 4x = 3y, (b) x = 4 or x2 + y2 = 4x + 4y, , Let a, b, c are the sides opposite to angles A,B,C respectively, a~b, a ~b S j a, b, c, in a AABC tan-------- =-------- cot — and------- =------- = -—■, 2, a+b, 2, sin A, sin B sin C, If a = 6, b - 3 and cos(A -B) = -|, , 116. Angle C is equal to, , «7, , OT7, , 4, , 2, , (c)^, (c)^, 4, , 117. Area of the triangle is equal to, (a) 8, (b) 9, (c) 10, (d) 11, 118. Value of sin A is equal to, , (b)42, , (a)-^, , ' ' V5, , (c) 3(x2 + y2) + 196 = 49 (x + y), , (d) None of the above, 112. If ADEF is equilateral, then ‘P, (a) coincides with incentre of AABC, (b) coincides with orthocentre of AABC, (c) lies on pedal A of ABC, (d) None of the above, , Passage III, (Q.Nos. 113 to 115), , In an acute angled AABC, let AD, BE and CF be the, perpendicular from A, B and C upon the opposite sides of the, triangle. (All symbols used have usual meaning in a triangle.), , 113. The ratio of the product of the side lengths of the ADEF, and AABC, is equal to, i, , (a), , 3(abcy, 4(a + b + c), , Passage V, (Q.Nos. 119 to 123), , When any two sides and one of the opposite acute angle are, given, under certain additional conditions two triangles are, possible. The case when two triangles are possible is called the, ambiguous case., In fact when any two sides and the angle opposite to one of, them are given either no triangle is possible or only one, triangle is possible or two triangles are possible., In the ambiguous case, let a, b and ZA are given and q, c2 are, two values of the third side c., , On the basis of above information, answer the following, questions, , 119. Two different triangles are possible when, (a)hsinA<a, (b) bsinA < a and b > a, (c) bsin A < a and b < a, (d) bsinA < a and a = b, 120. The difference between two values of c is, (b)/a2-b2), (a) 2-J(a2 - b2), , (c) cos A cosB cosC, , (c)2^a2 - b2sin2 A), , 114. The orthocentre of the AABC, is the, (a) centroid of the ADEF, (b) circum-centre of the ADEF, (c) incentre of the ADEF, (d) orthocentre of the ADEF, , 115. The circum-radius of the ADEF can be equal to, I X abc, , a, , “ 8A, , 4 sin A, , 121. The value of c2 - 2c, (a) 4a cos A, (c) 4a cos2 A, , (d) ^(a2 -82sin2A), cos2A + c2 is, , (b) 4a2 cos A, (d) 4a2 cos2 A, , 122. If ZA = 45° and in ambiguous case (a, b, A are given), , cv c2 are two values of c and if 0 be the angle between, the two positions of the ambiguous side c then cos 6 is, , (a)^T?, C + c2, , www.jeebooks.in, (c>7, 2, , A, B, c, (d)j cosec— cosec — cosec —, 8, 2, 2, 2, , i, , (c), , Cj + c2, , V^Z, , ^VClC2, , (Cj + c2), , (q + c2)

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www.jeebooks.in, ■, , Chap 03 Properties and Solutions of Triangles, , (m - l)(m + 3), , 123. If 2b = (m + l)a and cos A = 2, , m, , (Q.Nos. 127 to 131), AL, BM and CN are perpendicular from angular points of a, triangle ABC on the opposite sides BC,CA and AB, respectively. A is the area of triangle ABC, (r)and R are the, inradius and circumradius., , 1 < m < 3, then — is, c2, , 1, (m-1), _ 1__, (d) (m + 3) or, (m + 3), , 1, , (a) m or —, m, (c)(/n + l)or, , (b)(m -1) or, 1, (m+ 1), , Passage VH, , L where, , c, , . X, , 265, , 127. If perimeters of ALMN and AABC and A, and p, then the, value of — is, , Passage VI, (Q.Nos. 124 to 126), , Consider a triangle ABC, where x, y, z are the length of, perpendicular drawn from the vertices of the triangle to the, opposite sides a, b, c respectively. Let the letters R, r, S, A, denote the circumradius, inradius semi-perimeter and area of, the triangle respectively., m irbx cy az a2, 124. If — + — + — = —, cab, (a)R, , + b2 +c 2, —, then the value of k is, k, 3, , (c)2J?, , (b)S, , /, , X, , 1, 1, 1, 125. If cot A + cot B + cot C = k — + — + — , then the, z2, , value of k is, (a)B2, (c)A, 126 The value of cs*n, , + as^n, , x, bsinA + asinB., +--------------------is equal to, z, , (a)r, , (c)2, , (b)-, , 128. If areas of A's AMN, BNL and CLM are Ap A 2 and A 3, respectively, then the value ofA1 + A2 + A3 is, (a) A(2 + 2cosAcosBcosC), (b) A(2 + 2 sin A sin B sin C), (c) A(1 - 2 cos A cos B cos C), (d) A(1 -2sinAsinBsinC), , 129. If area of ALMN is A', then the value of — is, A, (a) 2sin Asin Bsin C, (b) 2 cos A cos B cosC, (c) sin A sin B sin C, (d) cos A cos B cos C, , 130. Radius of the circum circle of ALMN is, (a)2R, (b) R, , 0>)rR, (d) a2 + b2 + c2, +, , (a,i, , z x jR, , /, , (c) 2, , + cs^n, , y, , R, , (d) 4, , 131. If radius of the incircle of ALMN is r', then the value of, r'sec A sec B sec C is, (b)3R, (a)4R, (c)2R, (d)R, , (d)6, , g Properties and Solutions of Triangles Exercise 5:, Matching Type Questions, 132. Match the statement of Column I with values of, Column II., Column I, , (A) In a AABC, let Z.C = -,r = inradius, R=, 2, circumradius then 2(r + R), , Column II, , a+Z>+c, , (B) If Z, m, ware perpendicular drawn from, (q) a~b, the vertices of triangle having sides a, b, and cthen____________________, bl cm an\ n ,, 2R — + — + — + lab + 2bc + 2ca, a, bJ, N kc, , (C) In a AABC, R(b2sin2C + <?sin2B), , (r) a + b, , equals, , In a right angle triangle ABC, Z.C = p, , (s) abc, , v . n . A + B . (A - B), then 4R sin------- sm---------2, 2, , www.jeebooks.in

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www.jeebooks.in, 266, , Textbook of Trigonometry, , 133. Match the statement of Column I with the values of, Column II., , (C), , In a triangle ABC, then line joining the (r) 1, circumcentre to the incentre is parallel to, BC , then value of cos B + cosC is, , (D), , If in a AABC, a = 5, b = 4 and, 31, cos(/l - B) = —, then the third side cis, , Column II, , Column I, , (A) In a triangle ABC (c - a)2 = b2 - ac and (p) A =, cosB + sinC = —, 2, , (B), , equal to, , A, B,C are in A.P. andC = 3A, , (C) The length of the bisector of angle, , J3ca, , (q) B = 6(F, , 135. Match the statement of Column I with values of Column, , n, , (r) C = 9(F, , ,, , ,, , B =-------- and a = b, (c+ a), (D), , (s) 6, , Column II, , Column I, , 3, 1, I, +----- =, a+ b b+ c a+ b+ c, , (A), , In a AABC, if, 2a2 + b2 + c2 = 2ac + 2ab,, then, , (p), , AABC is equilateral, triangle, , (B), , In a AABC, if, a2 + b2 + c2 = J2b(c + a\, then, , (q), , AABC is right, angled triangle, , (C), , In a AABC, if, a2 + b2 + c2 = be + ca-jl,, then, , (r), , AABC is scalane, triangle, , (s), , AABC is scalane, right angled triangle, , (t), , Angles B, C, A are, in AP, , (s) A = B=C, = 60°, , 134. Match the statement of Column I with values of, Column II., Column I, , Column II, , (A), , In a AABC, (a + b + c)(b + c - a) = Xbc, (p) 3, where A e I, then greatest value of X is, , (B), , In a AABC, tan ,4 + tan B + tanC = 9. If (q) 9(3)1/3, tan2 A + tan2 B + tan2C = k, then least, value of k satisfying is, , g Properties and Solutions of Triangles Exercise 6:, Single Integer Answer Type Questions, 136. If in AABC, Z.C = ~,a = ^2 and b = -J2 + V2, then find, , the sum of digits in the measure of angle A (in degree)., 137. In the figure as shown, find the number of digits in the, length of AB., C, 195, , A, , y, , respectively. If O is the circum-centre of the AABC, then, f Area of AABC^, the value of, cot B cot C equals to, kArea of AOPQy, , 141. With usual notation in AABC, the numerical value of, f, a+b+c . a, b, c, — + — + — is, + r2 + r3 ) VI, , 125, , 3xv, , r2, , r3 y, , A(c2 - a 2), —, where AD is the, 3ca, median through A and AD ± AC, then the value of A is, , 142. In a AABC, cos A • cos C +, , c, 2, , 138. If A = —, B = — and C = — then in AABC, ~, 7, 7, 7, equals to, , + b2 +c2, R2, , 139. If A, B, C the angles of an acute angled AABC and, , D=, , 140. In a AABC, P and Q are the mid-point or AB and AC,, , (tanB + tan C)2, , tan2 A, , tan2 A, , tan2 B, , (tan A + tan C)2, , tan2 A, , tan2 C, , tan2 C, , (tan + tan B)2, , 143. In a triangle ABC, medians AD and CE are drawn. If, jr, , tt, , AD - 5, ZDAC = — and Z.ACE = —, then the area of the, 8, 4, 5a, triangle ABC is equal to —, then a + b is equal to, b, , www.jeebooks.in, then the least integral values of —— is, 1000, , 4, , •, , f, , /, , A, , 144. In AABC, — = - then the value of 16 E tan —, r, 2, 1, * 2*, , be., , must

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , A, 2^, a+b+c, , X, , Sa cos2, , 145. In a AABC, the maximum value of 120, , k, , must be, 146. The sides of triangle are three consecutive natural, numbers and its largest angle is twice the smaller one., The largest side of the triangle must be, 147. In AABC, ZC = 2ZA, and AC = 2BC, then the value of, a2 + b2 +c2, (where R is circumradius of triangle) is, R2, 148. If a, 5 and A are given in a triangle and cp c2 are the, , 267, , 149. In triangle ABC, a = 5, b = 4, c = 3. G is the centroid of, triangle. If R} be the circumradius of triangle GAB then, fl, , 2, , the value of — R. must be, 65, , 150. A triangle ABC is inscribed in a circle of radius 1 and, centre at O. The lines AO, BO, CO meet the opposite, sides at D, E, F. Then — + — + — is equal to, AD BE CF, 3 |, , I r3, , 151. In AABC, a > b > c and if-----------------------------= 8,, sin3 A +sin3 B + sin3 C, than the maximum value of a is, RS = 3 units and PQR = 609, then SP is, , possible values of the third side, and, cf +cj -20^2 cos A = Xa 2 cos2 A, then the value of X, , is, , Properties and Solutions of Triangles Exercise 7:, Subjective Type Questions, 153. In a AABC, the angles A and B are two values of 0, satisfying V3 cos 0 + sin 0 = K; where | K| < 2, then show, , triangles is obtuse angled., , = 3sin A • sin B • sin C, then find the value of determinant, a b c, , 154. If in an obtuse angled triangle the obtuse angle is — and, 4, the other two angles are equal to two values of0, satisfying a tan 0 + hsec0 = c, where |b| < yja2 + c2,, then find the value of a2 - c2., , 155. In a AABC, a, c, A are given and bv b2 are two values of, , third side b such that b2 = 2bv Then, the value of sin A., 156. If P is a point on the altitude AD of the AABC, such that, Z.CBP = —, then find the value of AP., 3, , 157. If R denotes circum-radius of AABC, evaluate, , 161. If in a AABC, sin3 A+sin3 B + sin3 C, , b2 -c 2, , 2aR, , 158. InAABC.A=-,b- c = 3>/3 cm and ar (AABC), 3, 9y/i, =---- cm 2 . Solve for side a., 2, , 159. Find the value of tan A, if area of AABC is a2 - (b - c)2., , b c a, , c a b, , 162. In a AABC, the side a, b and c are such that they are the, roots of x3 - llx2 + 38x - 40 = 0. Then, the value of, cos A cos B cos C, -+------- +------- ., a, b, c, 163. In a AABC the sides a, b and c are in AP. Evaluate, A, C, B, tan — + tan — cot —., 2, 2, 2, 3, 164. The sides of a A are in AP. and its area is - x (area of an, , equilateral triangle of the same perimeter). Find the ratio, of its sides., 165. If AD, BE and CF are the medians of a AABC, then, evaluate (AD2 + BE2 + CF2).(BC2 + CA2 + AB2)., , 166. AD is a median of the AABC. If AE and AF are medians, of the AABD and AADC respectively, and AD = mv, , www.jeebooks.in, 160. In a AABC, B = 90°, AC = h and the length of, , perpendicular from B to AC is p such that h = 4p. If, AB < BC, then measure ZC., , q2, , AE = m,, AF = m,, then find the value of —., 8

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www.jeebooks.in, 268, , Textbook of Trigonometry, , A, t. AB., A-B, ■ffi7, v = »tan-------PC., A, 70/. if, If X, ;; Yv, •tan, tan—, —, Y == tan, tan-------------- tan —C;, 2, 2, 2, 2, C-A, B ,, ~ . .., ., c, Z = tan -------- tan —, then find the value of, 2, 2, X+Y+Z + XYZ., , 168. Let AABC be equilateral on side BA produced, we choose, a point P such that A lies between P and B. We now, denote ‘a’ as the length of a side of AABC; r} as the, radius of incircle of APAC; and r2 as the radius of the, excircle of APBC with respect to side BC. Determine the, sum (f] + r2) as a function of ‘a alone., 169. A hexagon is inscribed in a circle of radius r. Two of its, sides have length 1, two have length 2 and the last two, have length 3. Prove that r is a root of the equation, 2r3 -7r-3 = 0., , 170. The base of a triangle is divided into three equal parts. If, 0lt02,03 be the angles subtended by these parts at the, vertex, then prove that, , (cot0j + cot02)(cot02 + cot03) = 4 cosec202, , 171. If the circum-radius of a A is, , 54, , and its sides are in, V1463, , 3, , GP with ratio -, then find the sides of the triangle., 172. Prove that a2 + b2 + c2 + 2abc < 2, where a, b, c are the, sides of triangle ABC such that a + b + c = 2., 173. Let points PVP2,P3 ...,Pn_x divides the side BC of a, AABC into n parts. Let r}, r2, r3,..., rn be the radii of, inscribed circles and let pv p2,..., pn be the radii of, excribed circles corresponding to vertex A for triangle, ABPp AP1P2, ..., APn _]C and let r and Pbe the, corresponding radii for the triangle ABC. Show that, , HH., , Ll = L, , P1P2'"Pn, , P, , 177. Let ABC be a A with altitudes hvh2,h3 and inradius r., _, , h, + r h, + r, +r, Prove that —----- + —------+ —------ >6., h\~r h2 -r h3 - r, tp*, a cos A + bcos B + c cos C a + b + c ., 178. If in a AABC,------------------------------- ---------- , then, a sin B + bsin C + csin A, 9R, prove that A is equilateral., , 179. In AABC, ‘h’ is the length of altitude drawn from vertex, A on the side BC. Prove that:, 2(b2 + c2)>4/i2 + a2. Also, discuss the case when, equality holds true., 180. Consider a AABC. A directly similar AAjBjCj is inscribed, in the AABC such that ApB] and Cj are the interior, points of the sides AC, AB and BC, respectively. Prove, thatArea(AA1B1C1) >1, , Area(AABC), , cosec22 A + cosec2 B + cosec2 C, , 181. Find the angle at the vertex of an isosceles triangle, having the maximum area for the given length 7’ of the, median to one of its equal sides., 182. Consider a AABC and points A] and Bx on side BC such, that ZBAAj = ZB] AC. If incircle of ABAA} and BjAC, touch the sides BAX and BjC at M and N respectively,, 1 =-----1 +----1 ., prove that ----1 +------BM MAX B{N NC, , 183. An equilateral triangle PQR is circumscribed about a, given AABC. Prove that the maximum area of APQR is, fl2, + ^2, b2 +cc2, 2A +-------- —----- . Where a, b, c are the sides of AABC, 2^3, and A is its area., 184. In a AABC, rA,rB, rc are the radii of the circles which, touch the incircle and the sides emanting from the, vertices A, B, C respectively. Prove that,, r, , 174. A polygon of n sides, inscribed in a circle, is such that its, sides subtend angles 2a, 4a,.... 2nd at the centre of the, circles. Prove that its area A], is to the area A2 of the, regular polygon of n sides inscribed in the same circle, as, sin nd : n sin a., 175. A,, A2, A3,.... A„ is a regular polygon of n sides, circumscribed about a circle of centre O and radius ‘a’. P, is any point distant ‘c’ from O. Show that the sum of the, squares of the perpendiculars from P on the sides of the, (, c2^, polygon is n a2 + — ., k, 2 >, , ■, , 185. Find the points inside a A from which the sum of the, squares of distance to the three sides is minimum. Also,, find the minimum values of the sum of squares of, distances., 186. In a scalene acute AABC, it is known, that line joining, circumcentre and orthocentre is parallel to BC. Prove, , that the angle A e —, — L, \3 2 ), , 187. Consider an acute angled AABC. Let AD, BE and CF be, the altitudes drawn from the vertices to the opposite, n, ,, EF FD DE R + r, sides. Prove that: — + — + — =------ ., a, b, c, R, , www.jeebooks.in, 176. Show that in any AABC, a3 cos3B + 3a2bcos(2B - A), + 3ab2 cos(B - 2A) + b3 cos 3A = c3, , 188. Two circle, the sum of whose radii is ‘a’ are placed in the, same plane with their distance ‘2a’ apart. An endless

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , I, , I, , string is fully stretched so as partly to surround the, circle and to cross between them. Prove that length of, string is (+ 273 k, , 190. Let Pbe the point inside the AABC. Such that, ZAPB = ZBPC = ZCPA. Prove that, , a2, , a3, , ao, , — T -------- T -------- “--------- —, , A, , A, , A, , A, , 2 +, b, l2 , _2, 2 +c, , — + 273A, where a, b, c, A, 2, are the sides and the area of AABC., , PA + PB + PC =, 189. If Ao is the area of A formed by joining the points of, contact of incircle with the sides of the given triangle, whose area is A. Similarly A1( A2 and A3 are the, corresponding area of the A formed by joining the points, of contact of excircles with the sides. Prove that, , 269, , 191. In an acute angled AABC, the points A', B' and C' are, , located such that A' is the point where altitude from A, on BC meets the outward facing semi-circle drawn on, BC as diameter, points B', C' are located similarly. Prove, that, {ar(BCA')}2 + {ar(CAB')}2 + {ar(ABC')}2 = {ar(ABC)2}, , H Properties and Solutions of Triangles Exercise 8:, KZu, , *, , ■■, , I, , I ■, , rx, , ", , 4 A if, , IP", , Questions Asked in Previous 10 Years' Exam, , (i) JEE Advanced & IIT-JEE, 192. In a AXYZ, let x, y, z be the lengths of sides opposite to, the angles X, Y, Z respectively and 2s = x + y + z. If, , -—— = -—- = -—- and area of incircle of the AXYZ is, 4, 3, 2, , —, then, 3, [More than one correct option 2016 Adv.], (a) area of the AXYZ is 676, 35 r, (b) the radius of circum-circle of the AXYZ is —V6, 6, , . , . X . Y . Z 4, (c) sin—sin—sm— = —, 2, 2, 2 35, (d) sin2, , x+y, 2, , 3, =—, 5, , 193. In a triangle, the sum of two sides is xand the product, of the same two sides is y. If x2 - c2 = y, where c is the, third side of the triangle, then the ratio of the in-radius, to the circum-radius of the triangle is, [Single correct option 2014 Adv.], (b) - -3y- (a), 2x(x + c), 2c(x+ c), (c) —-y-4x(x + c), , (d) ——, 4c(x + c), , 194. Consider a AABC and let a, b and c denote the lengths of, the sides opposite to vertices A, B and C, respectively., a = 6, b = 10 and the area of the triangle is 15 73. If Z.ACB, is obtuse and if r denotes the radius of the incircle of the, • triangle, then r2 is equal to......, [Intger Answer Type 2013 Adv.], , RP at N,L and M respectively, such that the lengths of, PN, QL and RM are consecutive even integers. Then,, possible length(s) of the side(s) of the triangle is (are), (b) 18, , (a) 16, , [More than one correct option 2012], (d) 22, (c) 20, , 7, 5, 196. If APQR is a triangle of area A with a = 2, b = - and c = -,, 2, 2, where a, b and c are the lengths of the sides of the, triangle opposite to the angles at P, Q and R, . , _, 2 sin P - sin 2P, respectively. Then,------------------- equals, 2 sin P + sin 2P, [More than one correct option 2012], , (a) —, 4A, , (b)£, 4A, , 3, , (45, , (c), , 4A, , (d), , s, , <4A, , 197. If the angles A, B and C of a triangle are in an arithmetic, progression and if a, & and c denote the lengths of the, sides opposite to A, B and C respectively, then the value, a, c, of the expression - sin 2 C + - sin 2A is, c, a [Single correct option 2010], ^3, r, (c) i, (d> -li, (b) Y, 198. Let ABC be a triangle such that Z.ACB = —.]fa,b and c, denote the lengths of the sides opposite to A, B and C,, respectively. Then, the value(s) of x for which, a-x2 +x + l,i> = x2-landc = 2x + lis(are), [Single correct option 2010], (a) -(2 + 73), (b) i + Ti, , www.jeebooks.in, 195. In a APQR, P is the largest angle and cos P = k Further, , in circle of the triangle touches the sides PQ, QR and RP, , (d) 473

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www.jeebooks.in, 270, 199., , Textbook of Trigonometry, , In a AABC with fixed base BC, the vertex A moves such, , that cos B + cos C = 4 sin2 —. If a, b and c denote the, 2, lengths of the sides of the triangle opposite to the angles, A, B and C respectively, then, [More than one correct option 2009], (a) b + c = 4a, (b) b + c = 2a, (c) locus of point A is an ellipse, (d) locus of point A is a pair of straight line, , 200. Let ABC and ABC' be two non-congruent triangles with, sides AB = 4, AC = AC'= 2^2 and ZB = 30°. The, absolute value of the difference between the areas of, these triangles is, [Integer Answer Type 2009], 201. A straight line through the vertex P of a APQR intersects, , the side QR at the point S and the circum-circle of the, APQR at the point T. If S is not the centre of the, circumcircle, then, [More than one correct option 2008], ,, , 1, , (a) — +, PS, ztl 1, (b) — +, PS, ., ', , 2, , 1, , (d) AAEF is isosceles, , triangle (in sq units) is, (a) 4^3, (b) 12-7^3, , [Single correct option 2006], , (c) 12 + 7-73, , (d) None of the above, 207. In a A ABC, among the following which one is true?, [Single correct option 2005], ^4, (a) (b + c) cos — = a sin, 2, , I 2 J, , B + C>, , (b) (b + c) cos, , . 2 J, , A, = a sin —, 2, , B-Cy, , (d) (b - c) cos — = a sin, , k, , 2, , 2 J, , (ii)JEE Main & AIEEE, , 4, , Passage, (Q.Nos. 202 to 204), Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They, , intersect at P and Q in the first and the fourth quadrants,, respectively. Tangents to the circle at P and Q intersect the, Z-axis at R and tangents to the parabola at P and Q intersect, the %-axis at S., [One correct option 2007], 202. The radius of the in-circle of APQR is, (a)4, (b) 3, , 208. Let a vertical tower AB have its end A on the level, ground. Let C be the mid-point of AB and P be a point on, the ground such that AP = 2AB. If ZBPC = 0, then tan (3, is equal to, [2017 JEE Main], , (a) 6-, , (b) 1, , (c) j, 9, , (d) |, 9, , 209. ABCD is a trapezium such that AB and CD are parallel, and BC ± CD, if Z ADB = 0, BC = p and CD = q, then AB, [2013 JEE Main], is equal to, (p2 + q2)sin0, p2 + <?2cos 0, (b), pcos 0+ qsin0, pcos 0+ <?sin 0, , (d)2, (c), , 203. The radius of the circum-circle of the APRS is, (a) 5, (b) 3^3, (c) 3V2, (d) 2^3, 204. The ratio of the areas of &PQS and APQR is, (a) 1: 41, (b) 1:2, , (c) 1 : 4, , A, 2bc, cos —, 2, b+c, , 206. In-radius of a circle which is inscribed in a isosceles, triangle one of whose angle is 2?t / 3 is 73, then area of, , 1, 2, — > , ..... —, ST y/QSxSR, , 1, , (b) AD =, , 4bc . A, sin —, b+c, 2, , (c) (b - c) cos, , (d — + — > —, PS ST QR, , (C)|, , (c) EF =, , — < , ----ST jQS x SR, , , . 1, 1, 4, (c) — + — < —, PS ST QR, 1, , (a) AE is HM of b and c, , (d) 1:8, , 205. Internal bisector of ZA of AABC meets side BC at D. A, line drawn through D perpendicular to AD intersects the, side AC at E and side AB at F. If a, b, c represent sides of, A ABC, then, [More than one correct option 2006], , p2 + g2, p2cos 0+ q2sin0, , (d), , (p2 + g2)sin0, (pcos 0+ <?sin0)2, , 210. For a regular polygon, let r and R be the radii of the, inscribed and the circumscribed circles. A false, statement among the following is, [2010 AIEEE], , (a) there is a regular polygon with — = R 2, r, 1, (b) there is a regular polygon with — =, R \2, T, , 2, , (c) there is a regular polygon with — = R 3, J3, (d) there is a regular polygon with — =, 2, R, , www.jeebooks.in

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 211. In a AABC, let ZC = —, if r is the in-radius and R is the, 2, circum-radius of the AABC, then 2(r + R) is equal to, [2005 AIEEE], , 212 If in A ABC, the altitudes from the vertices A, B and C on, , opposite sides are in HP, then sin A sin B and sin C are in, [2005 AIEEE], (b) AGP, , (a) HP, (c) AP, , (b) a + b + c, (d) b + c, , (a) c + a, (c) a + b, , 271, , (d) GP, , Answers, Exercise for Session 1, l.(a), 7-(d), 13. (c), 19-(c), , 2. (a), , 8- (c), 14. (a), 20. (a), , 3. (c), 9. (b), 15. (a), , 5.(c), , 4.(c), 10. (c), 16. (c), , U.(c), 17. (b), , 4.(d), 10. (b), 16. (c), , H. (d), 17- (a), , 6. (a), 12. (d), 18. (a), , Exercise for Session 2, l.(a), , 2. (a), , 7-(d), 13. (b), 19. (c), , 8. (c), 14. (b), 20. (a), , 3. (b), 9. (a), , 15. (d), , 5-(a), , 6. (c), 12. (c), 18. (C), , Exercise for Session 3, 1.1018.81 sq. cm, 26.(1), , 27.0, , 2. 2.5, , ,.sR, , 3.2 cm, , 7. 7:2, , 5.—, 2, , 6. 2a2b2c2, , 28. a2b2, , Exercise for Session 4, 1. a cot .4 2.-----a+ b, , 9J, 2, , 4. 4/3, , 10. acosec(A/2), , 1, , Exercise for Session 5, n»*, * au, 2. - cosec—, 2, n, , , a tn, l.-cot—, 2 2n, 6.7,3, , 4. cosec —, 9, , Exercise for Session 6, 1.2, , 2.2, , 4.-^a+ b, , 3.6, , 155., , 5. ntan^, , 9^7, 8?, , 158. (-763 - 2>/3), , Exercise for Session 7, l.-L 2. sin A + sin B + sin C, , 3.2, , 4.6, , 6. r = 1, , 9. (2-V3):2V3, , 7.4, , 29. (b), 30. (a), 28. (a), 26. (c), 27. (a), 25. (a), 34. (c), 36. (a), 32. (b), 35. (b), 33. (d), 31. (a), 40., (a), 42. (c), 37. (b), 38. (a), 39. (c), 41. (b), 47. (c), 48. (d), 45. (b), 46. (a), 44.(c), 43. (a), 53. (b), 54. (b), 50. (a), 52.(c), 49. (b), 51-(a), 60. (a), 58. (a), 59. (b), 55. (b), 56. (d), 57. (c), 64. (a), 65. (b), 66. (d), 63. (a), 62. (b), 61.(c), 69. (c), 67. (b), 68. (a), 72. (b,c,d), 70. (b,d), 71. (b,c,d), 75. (a,c), 74. (a,c,d), 73. (b,c), 78. (c.d), 77. (a,c,d), 76. (a,b), 81. (a,b,c,d), 80. (a,c), 79. (b.c), 84. (b,c), 83. (a,b,c,d), 82. (a,b,c,d), 87. (b.c), 86. (a,c,d), 85. (a,b), 90. (a,d), 88. (a,b), 89. (a,b), 91. (a,c), 96. (a), 97. (c), 94. (a), 95. (d), 92. (c), 93. (d), 102. (a) 103. (a), 100. (a), 101. (c), 98. (a), 99. (c), 107. (b) 108. (c) 109. (c), 106. (b), 104. (a) 105. (a), 114. (c) 115. (a,b,c,d), 112. (d), 113. (c), 110. (c) lll.(c), 120., (b) 121. (d), 119., (b), 118. (b), 116. (b) 117. (b), 124., (c), 125., (c), 126., (d) 127. (b), 122. (b) 123. (a), 131., (a), 130. (b), 128. (c) 129. (d), 132. (A) -> (r), (B) -> (p), C -> (s), (D) (q), 133. (A) -> (p, q, r), (B) -> (p. q, r), C -> (q. s), (D) —> q, 134. (A) -*(p), (B) -> (q), C -> (r), (D) -> (s), 135. (A) -> (p, t), (B) -> (q. s), C -»(q, r, t), 139. (2) 140. (4) 141.(4), 138. (7), 136. (9) 137. (3), 144.(8), 145. (9) 146. (6) 147. (8), 142. (2) 143. (8), 150. (2), 151.(2) 152. (2) 154. 2ac, 148. (4) 149. (5), , 161.(0), , 4. R tan a, , 5., , 2.05 sin 38°, , ■o, , 165. (3:4), , 167. (0), , 168., , 181. j —, , 185. A™, ran, , 192.(a,c,d), 196. (b,c,d), 200.4 sq units, 205. (a,b,c,d), 210. (a,b,d), , 193. (b), 197. (d), 201. (d), 206. (c), 211. (c), , 166. ml + m!2 - 2m2, , aV3, , ni.G.’,’), , X_ 4 (s - a) (s - b)I (s2- c)4js, , sin 42°, , Chapter Exercises, , 160.(15°), , *<), , 164. (3:5:7), , Exercise for Session 8, , 157. (sin (B-C)), , 156. 2csin, , ar: + b2' + c2'', , 194. (3), 198. (b), 202. (d), 207. (d), 212. (c), , 195. (a, b), 199. (b. c), 203. (b) 204. (c), 208. (c) 209. (a), , www.jeebooks.in, l-(a), 7.(b), , 2-(a), 8. (c, , 13. (c), 19. (b), , 14. (d), 20. (c), , 3-(c), 9. (a), 15. (a), 21. (b), , 4. (c), 10. (b), 16. (d), 22. (d), , 5.(c), , H. (b), 17. (a), 23. (b), , 6- (c), 12. (b), 18. (b), 24. (c)

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www.jeebooks.in, Solutions, , 4., , x, sin 105°, , 2, sin 45°, , A, , tA2, , 7, , 45 \ n, , c, , J3, , •, , 272(7? + 1), 272, , 2 cos 15°, z., , 1, , k, , 8, , sin 45°, , x, , y, x + k + y = 16, (t + 7) x t = 2 x(2 + 13), Also,, 1=3, =>, x(x + k) = 6 x (6 + 7), Now,, y(y + k) = 1 x 14, and, Solving, we get x = 10 - ^22, y = 6-722 andk = 2^22., , 2. We have, s =, No2,, , AP, . 2B, sin —, 3, , AP = 2c sin—, 3, , 6. Given, 2b - a + c, , 2sinB =sinA + sinC, A-C, B, B, A+C, cos, => 2 2sin—cos— =2sin, 2, 2, 2, 2, . B, , 1, , (90°^, , 2, , I 2 ), , =>, , . B, 1, Sin— = —r=, 2 272, , =>, , B, cos— =, 2, , *, , 2, , '1-1=4, 8 2V2, B, , 7?, , Hence, sin B = 2 sin — cos — =, 2, 2, 4, 7 BD AB, ' DC~AC, , 3. Using Napier’s Analogy, B, , => AD is the angle bisector, ., 2bc, A, AD =------ cos—, b+c, 2, , D, \a=4, , c=a, , A-B, tan', 2, , C, , D, , 8, , 196 + 225 -169, 420, 252 3, 0<A <—, 2., 420 5, 4, A = sm, 5, , C, , b=5, , P, , z2fi/;, *8/3, , 4, 5, Alternatively By using cosine rule in AABC, we get, b,2 + «2, c - a2, cos A =, 2bc, (14)2 + (15)2 — (13)2, 2(14) (15), , A‘, , c, , sin( 90° + k, 3, A, , a+b+c, = 21, 2, 2A, sinA = —, be, _______ 4, 2, V21-8-7-6 =~ 1415, 5, A = sin, , Hence,, , 5., , a -b, C, cos—, a+b, 2, , 1~(^~1)(2 + ^)=-L, 1 + (7? -1), ^5, A-B, , 2X5X3 |1 + cos A, 8, 2, 15, 4, , 1+1, 5, , 2, , 15 2, 4 X 75, , 3x75 x 75, 2x^5, , www.jeebooks.in, = 30°, , 2, A - B = 60°, , 375, , ■ ■, , 2

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , —?h, —h, 8. Slope of GC = -—, slope of AG = —, a, 2a, 3b, , 273, , 12. Let a and b the roots of x2 - lx + 8 = 0. Then, a + b = 7, ab = 8, , Also,, Now, , tan30° = -2a~ and a2 + b2 = 9, , C = 60°, 1 _ a2 + b2 -c,2‘, 2~, 2ab, , ab=(a + b)2 - 2ab - c2, , C(0, b), , c2 = (a + b}2 - 3ab = 49 - 24 = 25, c=5, , Thus,, , r Ra^c, - 8x5 _ 5, ” 2(a + b + c) " 2(7 + 5) " 3, , 13. Applying m-n theorem, , A, , ■, , (a. 0), , _, , =>■, , 5, , (BD + DC)cot — = DC cosB - BD cotC, 6, =>, (cotB - cotC)2 = 12, 14. A'C = b cosC, B'C = a cosC, , 3ab, 2(a2 + b2), a2 + b2^, , ~ab =, 2, , 35, , 7, , /it=4=. '3, , =>, , 3V3, k =3, , =>, , B, 71, , 9. 4sin A cosB = 1, so A and B cannot be —, 2, 71, , 71, , [as if B = —, then cos B = 0 and if A = —, tan A is not defined), 2, 2, „, , =>, , „, , 71, , 71, , C = -,B= — -A, 2, 2, f 71, 4sinA cos---- A = 1, , =>, , I2, , sin2 A = —, 4, , =>, , A=”, , =$■, , 6, , A'B'= c cosC, Similarly, A'C' = b cosB and B C' = a cos A, Now,, 4R = a cosA + b cosB + c cosC, =>, sinA sinB sinC = 1, , is not possible., 15. Let the sides of the triangle be 7,7 - d, 7 - 2d., , ., , 1, , => smA = 2, _ n, => B = —, 3, , So angles are in AP., , Since, the given angle is the greatest (being obtuse) angle of, the triangle, it is opposite to the greatest side of the triangle, and we have,, 72 = (7 - d)2 + (7 - 2d)2-2(7 - d)(7 - 2d)cosy, , 10. det A is twice the area of the triangle with vertices, (a, p) (b, q) (c, r) with sides 5, 6,7., A2 = $($ - a)(s - b) (s - c), , =>, , 16A2 = 18-8-6-4, , det B = (det A)2 = 4 A2 = 18 -8 -6 = 864, 11. Given that the circle passes through the circumcentre of, AABC. Therefore, the distance between circumcenter, incentre, = ^R2-2rR = r, =>, , r2 + 2rR - R2 = 0, \2, , =>, , n, , —, R, , C, , => This is only possible when ZA = ZB = ZC = —, so triangle, , J, ■, , A', , -r, , +2- -1 = 0, R, , 72 =2 x 72 - 42d + 5d2-2(72 - 21d + 2d2, =>, , d2 -9d + 14 = 0, , (d-7)(d-2) = 0, d =2, (d = 7 is not possible), Therefore, the sides of the triangle are 7 cm, 5 cm, 3 cm., 1, 2n, Area of the triangle A = - x 5 x 3 x sin—, =, , R=, , 4, , cm2 and the radius of the circumcircle, , 7x5x3, , W), , www.jeebooks.in, — = 72 - 1 and, R, r, ABC, cos A + cosB + cosC = 1 + — = 1 + 4sin—sin—sin—, R, 2, 2, 2 ., , =>, , 7x5x3 7-5, =------t=— =---- cm, 15V3, 3

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www.jeebooks.in, 279, , Chap 03 Properties and Solutions of Triangles, , =>, , a2 k, b2, c~2, —— + —• *" —7 = 0, X2 k2, k2, , or, , a2 + b2 -c2 = 0, , 42. v a(b - c) + b(c - a) + c(a - b) = 0, X=, , a(b - c)x2 + b(c - a)x + c(a-b) = 0, , a2 + b2=e2, , or, , Then, other root = 1, |X| = *^, * ‘ o(b — c), , =>The triangle is right angled., , ..2, , 41., , 18“, , 2, , — + — + — =---1!9!, , 3!7!, , 5!5!, , 111118“, , 1!9!, , 4., , 3’7!, , 4., , 5!5!, , 4---------- -------, , 3!7!, , 9!1!, , (2b)!, , :.a, b, c are in HP, , 10!, 10! 10!, 10!, 10!), ---- +----- + -— +------+----10!<l!9! 3!7!----5!5! 7!3!---- 9!l!j, , _1, , Then,, , 8“, (2b)!, , —, (10q + ,0c33 +, 10! “, , (v roots are equal), , ab - ac = ca - be, 2ac, b+, a+c, , (2b)!, , ----- 4-, , 1 is a root of the equation, , 111, ., - are m AP., a b c, s s 5, ., - are in AP., a’ b' c, s, , 10C5 +, , 1, S, b, , a, , 1,c, , 1 are in AP, , =>, , 2y9 _ 8“ _ 2,3a, 10! ’ (2b)! “ (2b)!, , =>, , =>, , a=3,b = 5, , Also,, , Multiplying in each by ---------- -------------- .then, (s - a)(s - b)(s - c), , 2b + a + c =>, , =>, , a, , 10 = 3 + c, , fl=3, b=5, c=7, , (s - b) (s - c) (s - c)(s - a) (s - fl)(s - b), are in HP., ’, ca, ’, ab, be, , •••(*), 2, , cosC =, , if A') ., B^ . ■ C, or sin2 — , sm — , sm 2, are in HP., 2, , a^+b2-c,2, 2ab, 9 + 25 - 49, , 1, , 30, , 2, , 43. Let a be the radius of the circle, then the ratio of the area of, regular polygons on n sides inscribed to circumscribing the, same circle is given by, , C = 120° and A, B<60°, , 1 ;, ^_2na, , tan A + tan B + tan C = tan A tan B tan C, => tan A + tanB - -73 = -^tanAtanB, => tan A + tanB, , =>, , tanA + tanB > 0, 73(1 - tanAtanB) > 0, , =>, , tanAtanB < 1, , (ii), , =>, , 71, V3, — or, n, 2, , 6, , V, , sinA;, , (if A =90°), , Then, sinB + —— > 2, sinB, , 2a/X, , 3X2 - 10X + 3 £ 0, , and sinC + —— >2, sinC, , (3X - 1)(X -3) £ 0, , 3X -1 < 0, , n, , 44. fl[ sinA +1 + —— I, , V sinA + —— >2, sinA, , X-3<0, , 4, , rt=6, , (iii), , 2, , =>, , co, , or, , From Eqs. (i) and (ii), we get, ^(1-tanAtanB) ^(tan?ttanBy, , Let tan A tan b = X, y/3(l — X), , 3, , ,2, , => cos', , s2, , = -^3 tan A tanB, , tan A + tanB = V3(l - tanAtanB), Also,, , c, , be, ca, ab, are in AP., (s - b) (s - c)’ (s - c)(s - a)’ (s - a)(s - b), , c =7, , Also,, , b, , [from Eq. (iii)], , | sin A + —— +11| sinB + —— +1, \, sinA, A, sinB, , sinC + —-— + 11 > 27, sinC, J, , www.jeebooks.in, n| sin A + —-— + 1 | > 27, sin A, , X£3, , tanAtanb < 3, , or, , n, , sinA + sinA + 1, , sinA, , >27

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www.jeebooks.in, 280, , Textbook of Trigonometry, , 45. From the given information it is evident that distance between, I (incentre) and O (circumcentre) should be equal to inradius of, triangle., A, , A, , B, , C, , AO = R, ZOAI=2, C-B, 2, , --cl, 2, J, , Then, other function is F(x) = -2x + 8 = g(x) (say), c =/(I) + g(l) = 4 + 6 = 10, Now, x2 + y2 - xy = 10, 2, , x-y, ___ 2>, , r2 = R2 + 16J?2sin2—sin2—, 2, 2, 2, B, C, | C-B, -8k sin—sin—cos, 2, 2, ' 2, , c-bY, , 2, 2/, B CI, B C, => r = R (1 + 8sin—sin— 2sin—sin---- co;, 2, 2(22, 2, •2, , B, , C(, , DB, , c.C, , JDB, , CC^, , B+C, B, C, = /? 1 -8sin—sin—co:, V, 2, 2, . 2, ,2, , = K2 1 -8- — = R2 - 2rR, 4K., z, , /10, , 2, , =1, , is an ellipse whose cente (0, 0)., Maximum distance from origin on any point on ellipse = Semi, major axis = V10, , Then,, , a = r2 =10, , :., , a = c = 10, sinA : sinC = 1:1, , 48. Let a be the length of each side of the equilateral triangle ABC., , Then r, the radius of the in-circle =, , (the altitude., , median and the angle bisector of angle A), 1 [2 Q, a, 2, => r = -da, ----- -- —7=, 3V, 4 2V3, , ABC, = F■212 1 -8sin— sin—sin—, I, 2, 2, 2., , \22, , +/, , r = V10, , = R (1 + 8sin—sin— sin—sin----- cos—cos—, 2, 2<, "2, ”2, 2, 2>, , z, , Tao?, , 2 J., , Zl, B, C,, „B, B, C, C, = R■ 2 1 + 8 sin—sin— sin—sin---- cos—cos—, t, 2 2, 2, 2, 2, 2, 2,, , -, , [1.2]->[4,6], F(l) = 4, A + B =4, and, F(2)=6, A =2, B = 2, Then, one function is F(x) = 2x + 2 = /(x) (say), F(1) = 6=>A + B=6, A - -2, B =8, F(2) = 4 => 2A + B = 4, , C-B, Now, IO2 = OA2 + Al2 -2(OA)(A/)cos, 2, =>, , j, , 47. Let linear function is F(x) = Ax + B, , If ‘f and 'R' be the inradius and circumradius respectively,, then, A, „ . B . C, Al =r cosec — = 4Ksin—sin—, 2, 2, 2, and, , => sin A + sinB + sinC = 2sinA(sin A + sinB + sinC), • . = -1, =>, sinA, 2, =>, A = 30°, , A, , \, , I r ), ( r}, - +2 - -1 =0, , \RJ, , =>, , \RJ, r -2 ± 74 + 4, R, 2, , I, I, I, , ^2 - 1, as — > 0, R, I, I, I, , 1 + - = y/2, R, cosA + cosB + cosC = V2, , ;O, , s,, , I, , Hence, (b) is the correct answer, , 46. a + b + c *, , 1, I, I, , B-, , 3, , a + b + c = 2(sinA + sinB + sinC), _ sinB sinC, „, 1 +------ +------- = 2(sin A + sinB + sinC), sinA sinA, , C, , P, , Area of the square PQRS inscribed in this circle, = PQ2=OP2 + OQ2, , www.jeebooks.in, b, (using —-— =, = —), sin A sinB sinC, , 2, , *, , <2, , a, , = 2rz = 2 X-------= —, , 4X3, , 6

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www.jeebooks.in, 282, , Textbook of Trigonometry, , a, b2-c\ a b2-c2 b, abc 2, 2c, 2ac, :2c, b, a, 2c 2(AD), sin A, 1, 2sinC 2sinC, sinA -1, In AACD, cosB =, 2sinC, AD ., b, 1, — = sinC => — =----b, AD sinC, =>, 2cosBsinC =sinA -1, sin(B, + C) - sin(B - C) = sin A - 1, =>, sin A - sin(B - C) = sin A - 1, sin(B - C) = 1, , Since b is real, we have, =>, (x + I)2 - 4(2x + 1) > 0, , The least integral value of x is 7., , 56. Let r = 1 cm, a = BC = BD + DE + EC, = BD + C|C2 + BD, = 2BD + C,C2, = 2rcot30° + 2r, = 2(43 + l)cm, , Area of triangle, , B-C = 2, n, B = C+ - =23 + 90 = 113°, 2, ZB = 113°, 2A-B, 31, 1 - tan2------54. We have, — = cos(A - B) =----------- ., _, 32, l + tan2±^, 2, A-B, 63 tan2, =>, =1, 2, A-B, 1, tan------463, 2, A-B a-b, C, Now,, tan------cot —, 2, 2, a+b, , ABC = —a2 = — • 4(43 + l)2 cm,2‘, 4, 4, „„,2, = 45(4 + 243) cm, , = (6 + 445) ^2, cm, A, , 1, 5-4, C, cot —, 2, *63 - 5 + 4, , c, tan- =, Also, , », , <r, B, , r, , I, , /<30°V,, , 9, , D, , \r, ■, , hE, , 30°<A C, , 57. We have, 2b = a + c, , cosC =-------- W, 1 + tan2, , 1-^, 1, _, 81 _ 18, " 1 63 144 8, 81, c2 = a2 + b2 - 2abcosC, = 25 + 16-2-5-4-, , r, , I, I, , ^63, , 2, , x2 - 6x - 3 > 0, , =>, , 1, = 36, 8, , Hence c = 6, , 55. Using cosine rule, we get, x2 = (x + I)2 + b2 - 2(x + l)b cos3, 0 = 2x + 1 + b2 — (x + l)b, , 2sinB =sinA + sinC, B, B „. A+C, A-C, 4sin—cos— =2sin, ~^..L----------- COS----------2, 2, 2, 2, A, B A-C, = 2 cos — cos------2, 2, „ . B, A-C, 2 sin— = cos------2, 2, n A+C, A-C, i.e„, 2 cos------- = cos-------2, 2, A-C, A+C, Now, cos A + cosC = 2 cos --------- cos, 2, 2, A+C, n, A + C(, = 2 cos------- 2 cos, 2, 2 k, A+C, = 4cos2, 2, , •W, , [using Eq. (i).], .(ii), , www.jeebooks.in, b2 — (x + l)b + 2x + 1 = 0, , 4(1 - cosA)(l - cosC)= 42sin2y-2sinZy

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www.jeebooks.in, 284, , Textbook of Trigonometry, sinA, , y, , sinB, , x, , sin A - sinB, sin A + sinB, , y-x, y+x, , V3, , BC = ^AC2 - AB 2 = 7144 - 48, , [By Componendo and Dividendo], B, , = 796 =4^6, y, , Area = AB x BC = 473 x 476 = 4872 sq m., D, , 2(b/a) _p+b, 65. tana = - tan2a =, a, l-(b/a)', , x, , 45X, , T, , A, , C, , A + B . A-B, 2 cos------- sin------' 2 _y~*, 2, A+B, A-B y + x, 2sin------ cos------2, 2, A-B _ y -*, tan------2, y+x, y-x, y+x, , p, , I, b, , o-, , [As A + B = 90°], , t, 1, , (given), , x: y = (1 - t): (1 + t), , =>, , s(s - a)(s - b)(s - c) = 2s(s - a), A2 = 2s(s - a), A^, , 2(s -a), , s2, , s, , =1, , V^b2, , p+b, a, , b(a2 4- b2), 2ba2 - a2b + b3, -----------------=, P, ^, P, =, (a2-b2), a -b, , ..TO, , 66. x = /icot3a, (x+100) = h cot 2a, (x + 300) = h cot a, , 63. BD = (s - b),CD = (s -c)=$(s - b)(s - c) = 2, =>, , 2ba, , =>, , y 1+t, again by Componendo and Dividendo, — =-----x 1 -t, -1, , 1, , a, , ...(ii), ...(iii), , (radius of incircle of triangle ABC), , a, , - = constant,, s, , h, , 1, Now, A = -aHa, where lHa' is the distance of‘A’ from BC., 2, A 1£^ = 1 ’, =>, 2 s, s, 2s, =», Ha = — = constant, a, => Locus of ‘A’ will be a straight line parallel to side BC., 64. Let AE is vertical lamp-post., Given, AE = 12 m, D, , tan 45° = —, AC, AC = AE = 12 m, Un60°=^, AB, , C, , 2a, , a, I—200, A, , F, , 100, , B, , 3a, , 4C, , x, , From Eqs. (i) and (ii), - 100 = h(cot3a - cot 2a), From Eqs. (ii) and (iii), -200 = /i(cot2a - cot a), sin a, sin a, 100 = h\, and 200 = /i, sin2asina,, sin3asin2a?, sin3a 200, sin3a, or, ------- =----- => ------- = 2, sin a-- 100----- sin a, 3sina - 4sin3 a - 2sina = 0, =>, , 4sin3a-sina = 0 => sina = 0, , or, , sin a = — = sin —, 4, \6j, , • 2, , 1, , • 2^, , n, a=—, , www.jeebooks.in, 6, , Hence, H = 200sin- = 200— = 1Oo73, 3, 2, , [form Eq. (i)]

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www.jeebooks.in, 286, , Textbook of Trigonometry, , 73. (a) a22 -(c-b)2 = 4bc, , 75., , a = 10V3, s, , => a2 - (c2 + b2 - 2bc) = 4bc, , r = — = A and, s, , a2 = b2 + c2 + 2bc, , D, , r = ($ - b) tan—, , -2bc b2 + c2 - a2, 2bc, 2bc, => cosB = -1 which is not possible., (b) 2bzsinC cosC + 2c2sinB cosB = ab, ■, , 1, , 1, , = 10, , —, , s - = 3 => b=7, a + c = 13, ac = 40, A, , =>, , =>, , (b2c cosC + c2b cosB), ,, ---------------------------= a b, R, be, —(bcosC + ccosB) = ab, , •••(>), ••(ii), b, , c,, , 60°, B, , abc ,, c, 1, ---- = ab,=$ — = R, 2R 2, , =>, , sinC = - => C = —, 2, 6, (c) For a = 3, b = 5, c = 7, we have, a„2 J., + 0h2 — X-c 2, cosC =, 2ab, , From Eq. (ii), a2 - 13a + 40 = 0, , =>, , -15, 30, , 9 + 25-49, 2X3X5, , -1, 2, , 2, 74. We have, —-— = —— = —-—, sin A sinB sin30°, So,, 2sinA=sinB, Hence, option (a) is correct., Also, c2 = 4 + 1 - 4cos30°, A, , A, a., //, , 30/, , a=1, , C, , — =2R, sinB, 7, 7, 2R = ~ => B = Z=, , 73, , 73, , 2, , 76. cosC = - =, 2, , Also,, , O,, , 8, , =>, a = 5, a = 8, =>, c = 8 and c = 5., Hence, a = 8, b = 7 and c = 5 (As a > b > c), A, A, A, ...., 1 1, => r, =------ ,r2=----- -,r3 =----- will give —1 : —1 A, 12, s -a, s -b, s -c, 12 13, 12 ’ 15, , Also,, , C=^, 3, A-C, A+C, = cos, (d) As cos, 2, 2, „ . A . C, =>, ‘ 2sin—, 2 sin—, 2 = 0 which is not possible in triangle., , = 5-275, , ‘C, , a, , 16 + 64 - c2, 64, =, c2 48 => c = 475, a, sin A, 4, sin A, , c, sinC, 4j3, , 7|, , 2, A =30°, As, ZC = 60°, ZA =30°, ZB =90°, 1, -ac, A, ac, 2, Now, r ~ — =, a+b+c, s, a+b+c, 2, 4(475), , c = ^5 - 2^3, , 4 + 475 + 8, , As c > a => ZA < ZC and ZB is obtuse., 1 . n 1 1, A =-xlx2x- = Also,, 2, 2 2, , 1675, , =>, , So,, , 4A, 1 X2XC, , = c = AB, 4A, Hence, option (d) is correct., , 12 + 4yfi, , 475, 3 + 75, , Now, verify alternatives., (c)r =, , 475, 3 + 75, 8, , www.jeebooks.in, (d) T the length of internal angle bisector of ZC is, , ■v3, , Hence, options a and b are correct.

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , 77 p, b, b, n. R =-------= - cosec a, 2sin a 2, , 287, , a2 + b2 -c2 = -ab, , A = -b2sin(180° -2a) =-bzsin2a, 2, 2, A, , cosC = - 2, ZC = 120°, and 72(sinA + cosA) =, , fl, , ^■V2sin[ A + -|=73, \, , b, (O, RS, , 4/, , 7t 1 V3, A+— =, 2, V, , ', , ’■••••.fl, , 71, , B, , 71, , A+—=, 4 3, , arx/, ~7c, , ,g, , A = ~ possible, ., a, a, and r = 4R sin — sin—■ sin(90° - a), 2, 2, 2b, . 9a, sin a, =------ sm — cos a------sin a, 2, sin a, b(l - cosa)sin2a, bsin2a, 2(1 +cos2 a), 2(1 + cos a), , and, , 01 =^R2 -2Rr) = R, , (d), sin A + sinB = -----2, cosAcosB = — = sin Asin B, 4, cosAcosB -sinAsinB = 0, cos(A + B) = 0, , and, , VI r), , A+ B=—, 2, , = Rjl -4cosa + 4cos2 a = R(2cosa -1), , f(, , B = —- A, 2, , ,a, , = R<2 2cos--l, \, 2, f, , From Eq. (i),, , 9a, , = R 4 cos2---- 3, k, 2, , sinA + cosA =, sin^ A +, , R 4cos3 — -3cos—, I, 2, 2., a, cos—, 2, _ (3aA, R cos —, , k2 J=, a, , cos—, 2, , Ji + 1, 2, , =, , 2^2, =>, , Ht), , 2sinacoi, , —0), , 4, , . 5n, = sm—, 12, , 12, , A-i, , =>, , 6, , a, , B= —, 3, , 2., , Then, C = y, — possible., 78. (a) Since, tan A + tanB + tanC = tan A tan B tan C, , But here, tan A + tanB + tan C = 0, impossible, (b), sin A sinB sinC, 2, 3, 7, a, b, c, ———, or, 2 3 7, a+b c, =>, 5, 7, a+b, or, c, , 7, , 79. In AABC, b + c-a> 0, c + a-b> 0, a + b - c> 0, so, , (b + c - a) + (c + a - b)(a + b - c), 3, > {(b + c - a)(c + a - b)(a + b - c)}1/3, , p3 > 27 (b + c - a) (c + a - b) (a + b - c), , =>, , Also,, , (a + b + c) + (b + c ~ a) + (c + a - b) + (a + b - c), , 4, > {(a + b + c)(b + c - a)(c + a - b)(a + b - c)}1M, , ^>(16A)w, , www.jeebooks.in, (c), , a + b < c impossible,, (a + b)2 = c2 + ab, , P > 4A1/4 orP4 =256A

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www.jeebooks.in, 290, , ■, , Textbook of Trigonometry, , i-, , 4, 93. $! : cos(A - B) = |, , 4b2c2, s(s -a), Also, (AD)2 = -------- r X, be, (b + c)2, , (A-B, 1 - tan2 ( 2, , bc2s(2s - 2a), (b + c)2, , 4, 5, , (A-B, 1 4- tan2!, , + b + c)(b +--------c - a), _ bc._(a--------------(b + c)2, , I 2, , 2 tan2, , 'A-B'', 2 j, , ■, , Using, tan, , bc{(b 4- c)~ - a, (b + c)2, A, , 1, 9, 1, 3, , 2, 'A-B>, tan, 2 >, , [as a > b => A > B], A, , C, a-b, cot—, we get, 2, a+b, , A-B, , 2, , s, , 1 6-3, C, C ,, - =------ cot— => cot — - 1, 3 6+3, 2, 2, =>, ZC = 90° => Statement I is false., S2: Using sine law in AABC, we get, a " _ ——c —, sinA sinC, , b, , c, f, I, I, , B, , C, , D, , 97. IC1, = 2, ft, , a _ i/a2 4- b.2, . it, sin A, sin—, 2, =>, , 6, , I—, , 2, , ------ = V45 => sin A = —f=, sin A-----------------------V5, Statement II is true., , 94. Statement H can be proved (by using A 4- B 4- C = n) to be, true, (conditional identities) From which we get, 3-2(sinzA + sin2B + sin2C)=-l - 4cosAcosBcosC, , => 3 -2(2) = -1 - 4cosAcosBcosC, => cosAcosBcosC = 0, => one of the angles A, B, C is equal to 90*., 95. v A + C = 180°, B 4 D = 180°, , ‘, , :. cos A = - cosC, cosB = - cosD, cos A 4- cosB 4- cosC 4- cosD, = (cosA 4- cosC) 4- (cosB 4- cosD), = 0+0, =0, EcosA = 0, sin A =sinC, and, and, sinB =sinD, EsinA = sinA sinB + sinC + sinD, Then,, = 2(sinC + sinD) 0, , 96. Area of AABC = Area of AABD + Area of AACD, ifrcsinA, 1, . . = —, 1 c • AD sin^, (—, A\j 4- y, 1 b-ADsin^y^, . ., 2, 2, , , ., bcsmA, , 2bcco, , BICIj, is cyclic, Quadrilateral BP■ PC = IP-I}P, Hence, (c) is the correct answer., 98. Statement II is true., Statement I tan A = tanB = tan C, A = B = C ie, a = b -c, ri=r2=r3, r» + r2 + r3 _, 3-5r, r, A, a 4- b 4- c n, s-a, = 3----------- = 9, = 3A, b 4- c - a, s, , Hence, (a) is the correct answer., 99. Statement II is False. Because ifp=2,q = 4, r=6, then, p: q : r = 1:2 :3 but p 1, For statement I, let tan A = k, tanB = 2k, tanC = 3k, then from, tan A + tanB + tanC= tanAtanBtanC (in a triangle) we get, 6k = 6k3 =>k = 0,1, -1 but k = 0, -1 is not possible. So k = 1 =>, , www.jeebooks.in, b+c, , tan A = 1 => A = 45°. So, statement I is correct.

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www.jeebooks.in, =, , Chap 03 Properties and Solutions of Triangles, , X = tan^ B -C j, , A, •tan—, 2, ,, b-c, AA, AA b - c, X =------- cot — • tan— =------b+c, 22, 2 b+c, , So,, , ■, , 2 J, , Similarly,, , T=, , Hence — 4- — = 9(f---- -,, 2!, 2, 2, = cos ^90° - —] = sin —, and cos ------ T ——, 2, 2 ., 2, I, 2, , Next we apply the addition formula for the cosine;, oq, a2, . oq . a2, . cu, cos — • cos— - sm — • sin — = sin —, 2, 2, 2, , a -b Z=^, a + b’, c+a, , Now,, , X + Y + Z + XYZ, b-c -----a -b +------c — a+, =------+, b+c q+ b c+a, , i, , a-b, , c-a, a\r, , c + aj, , _(b-c)(a + b)(c + q) + (q - b)(b + c)(c 4- q) + (c -a)(b 4- c)(q 4- b), (b + c)(a + b)(c + q), + (b - c)(q - b)(c - a), (b + c)(q 4- b)(c + a), After multiplication and solving, we get, X + Y + Z + XYZ =, , - -= 0, (b + c)(b + q)(c + q), , 168. Looking the figure, we see that ZT}OXR = 60° and it is the, supplement of Z7JAJ? = 120° {as an exterior angle for AABC}, , Hence,, , ZAQB = 30°, , Ti, , c, , 8 r2, , = fj tan30° + a + r2 tan30° = -r*, , ,, . Otj, 1/2!, oq, where, sin — = — ,cos—2, r, 2, , [from (A)], , 2r, , . a2 1, a2, sm— = cos— = - ------[from (B)], 2 r, 2, r, . a3 3/2, sm— =---[from (C)], 2, r, We substitute the expressions into Eq. (i) and obtain, after, multiplying both sides by 2r2,, , and square, obtaining., , + q., , T/T^T^C + CT/, = CR + CS = (q - RA) + (a-SB), = 2a -, , c, , A, , Now, write it in the form;, V(4r2 - l)(r2 - 1) =(3r + 1)., , Similarly, we obtain ZBOyS = 30°, Since tangents drawn to a circle from external points are equal,, we have, T|T2 = TyA + AB + BT2 = RA + SB + AB, , and, , 112, , -1 -1 =3r, , I, , P, , 303, , (4r2 — l)(r2 -l) = 9r2 + 6r + 1, , which is equivalent to,, r(2r3 - Ir - 3) = 0, since r # 0 we have,, 2r3 -7r-3 = 0., , 170. Let, BP=PQ=QC=x, Also, let, A, , ri +, , ^3 ), , Since, common external tangents to two circles are equal,, Hence,, , r. + r3, 0i + r2), — 3 + a = 2q V3, , £, , Hence, we find that r2 + r2 = ~~’, , B, , x, , P, r, , x, , Q, w, , x, , C, , ZBAP = Q1, ZPAQ = 02, ZQAC = 03, and let ZAQC = 0, , 169. Equal chords subtend equal angles at the centre angles at the, centre of circle; if each of sides of length i subtends an angles, a( (i = 1,2,3) at the centre of the given circle, then, 2aj + 2a2 + 2a3 =360°, , Applying m: n rule in AABC,, , (2x + x)cot0 = 2xcot(0j + 02)-xcot03, 3 cot 0 = 2 cot(0j + 0,) - cot Qj, , G), , www.jeebooks.in

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , cosP =, , b2 + c2 - a 2, , 198. Using, cosC =, , a. b, A2 —cJ, a„2 +, , 2ab, , 2bc, 1 _ (2n + 4)2 4- (2n + 2)2 - (2n + 6)2, 3”, 2(2n + 4) (2n + 2), , v cosP =, =>, , 4n2 -16, 1, 8(n + 1) (n'+ 2) ~ 3I, , =>, , (n~2) =1, 2(n + 1) 3, , 311, , A, , 1, 3, , given, , n2 - 4, _1, 2(n + 1) (n + 2) ” 3, , c, , a=x2+x+1 B, , 73 _(x2 + x+l)2 + (x2-l)2-(2x+l)z, , 2(xz + x+l)(x2-l), , 2, , => (x + 2)(x + l)(x-l)x + (x2-l)z = V3(x2 + x + l)(x2 —1), , 3n - 6 = 2n + 2 => n = 8, Sides are (2n + 2), (2n + 4), (2n + 6), i.e. 18,20,22., , => x2 + 2x + (x2-1) = 73(x2 + x+1), , 196. If AABC has sides a, b, , (2 - 73) x2 + (2 - 73) x-(73 + 1) = 0, , =>, , x = -(2 + 73) and x = 1 + 73, , =>, , x = -(2 + ^), , But, , => c is negative., x = 1 + 73 is the only solution., Then, tan (A/2), , 4, , (s~b)(s -a), , _, , 199. Given, cosB + cosC = 4sin2 —, 2, , A, , s (s - a), , y, , A, , n 7 . 5, 2+-+a+b + c, ?, where, s =------------ =----- - — 2=4, 2, 2, 2 sin P -sin 2 P •_ 2 sin P (1 - cos P), 2 sin P +sin 2 P 2 sin P (1 + cos P), , fl<i=a=^C, , 2sinz(P/2), 2/n.«x, =------------ L = tan2 (P/2), 2 cos2 (P/2), , Fixed base, , P, , => 2cos, , b = 7/2, , c = 5/2, Q-, , =3, , B-C, . A, -2sin— =0, =>2sin— cos, 2, 2, 2, , R, , a=2, , cos, , (s-b)(s-c) y{s-b)(s-c), s (s - d), , As, , (s - b) (s - c), 2, , 2, , [(S-»2(i-c)2]_, A2, , 4-__ 2., , 4-2., , A2, , ■(a', , B-C ,, , 2 J, , -2 cos, , B+C, , 2, , =0, , • — * 0n, sin, 2, B, C , . B . C n, -cos—cos—+ 3sin—sin— = 0, 2, 2, 2, 2, B C 1, tan—tan— = 22 3, l(s-a)(s-c) (s-b)(s-a) _ 1, , y, , 197. Since, A, B, C are in AP., , =>, , B-C, B+C, A . 2A, = 4 sin —, cos, 2, , 2 ., 2, , s(s-c), , 3, , s-a 1, ------- = —, s, 3, , 2B = A + C i.e. ZB =60°, a, c, -(2sinCcosC) + - (2sinAcosA), c, a, = 2fc(acosC + c cos A), a, b, c, using,------ =------ =-----sin A sinB sinC, , s(s-b), , 2s = 3a, , b + c = 2a, , Locus of A is an ellipse., 1, , 2^2, 4, 200. In AABC, by sine rule, —— =, sin A sin30° sinC, =>, C = 45°, C = 135°, When, C = 45° => A = 180°-(45°+30°) = 105°, When, C - 135° => A = 180°-(135°+ 30°) = 15°, , www.jeebooks.in, = 2fc(b) = 2sinB, , = 73, , k., , [using b = acosC + ccosA]

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www.jeebooks.in, 312, , Textbook of Trigonometry, , V :, /, /, , C, , I, I, , \45* :, , 4, 2, 2, >— => >, QR, -JSQ-SR, y/SQSR, QR, _4_, 2, 1, 1, —+—>, ! >QR, y/QS-SR, PS ST, A, 202. Radius of in-circle is, r = —, 1, , A, , 4, ■, , \l35#, , 30', , B, , c, , Area of AABC = -ABx AC sin A = - x 4 x2^2 sin (105°), 2, 2, 73 + 1, = 2(73 + 1) sq units, = 472 x, 272, Area of AABC* = - AB x AC sin A, 2, = -x4x272 sin (15°) = 2(73 -l)sq units, 2, Difference of areas of triangle, =|2(73 +1)-2(73 -1)|= 4 sq units, , Alternate Method, C'_____ _, , $, , A = 1672, _ 672 + 6y/2 + 472 _, 5~, 2, , Since,, , Now,, , 1672, r" 872 = 2, , i, , 203. Equation of circum-circle of APRS is, (x+l)(x-9)+yz + Xy = 0, It will pass through (1,272), then -16 + 8 + A. -272 = 0, , A = -4==272, 272, /. Equation of circum-circle is x2 + y2 -8x + 272y -9 = 0, =>, , Hence, its radius is 3^3., , D, b\2, , Alternate Solution, , 2^2, , 30°, , ZPSR=B =*> Sin0=—;=, 273, , Let, B, , =>, , • 0n = —, PR, sin, 2R, PR =6^2=2R- sin0, , =>, , P=3y^, , 204. Coordinates of Pand Q are (1,272) and (1, -272)., , Here, AD = 2, DC = 2, Difference of areas of AABC and AABC, = AreaofAACCz, , Now, PQ = 7(472)z + 02 = 472, Area of A PQR = | • 472 8 = 1672 sq units, , = -ADxCC =-x2x4 = 4sq units, 2, 2, , 201. Let a straight line through the vertex P of a given A PQR, intersects the side QR at the point S and the circum-circle of, A PQR at the point T., Points P, Q, R, T are concyclic, then PS-ST = QS- SR, , Area ofAPQS = |-4^-2 = 472 sq units, Y-, , >|<(1?2j2), X', , __ S/ U0), , (-3,0), , R., , O, , <<-i. oy, , X, , (9,0), /Q\, , (1,-2-^y, Now,, , and, , Also,, , PS + ST, > J PS-ST, 2, 1, 1, 2, —+—>, PS ST JPS ST, SQ + QR, >y/SQ-SR, 2, ^>y/SQ-SR, , [v AM > GM], 2___, y/QS-SR, , Y‘, , Ratio of areas of APQS and APQR is 1 : 4., , 205. Since, AABC = AABD + AACD, =>, , f, , - be sin A = - c AD sin — + - b AD sin —, 2, 2, 2 2, 2, ., 2bc, A, AD =------ cos —, b+c, 2, , www.jeebooks.in, =>

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www.jeebooks.in, Chap 03 Properties and Solutions of Triangles, , A, , 2 sin, , A : A, 2 : 2, , b, , b+c, a, , B, , '' D, —a, b— c, , Also,, F, , Again,, , I 2, , A, cos —, 2, , r, , EF = ED + DF^2DE = 2AD tan 2, n2bc, AA, A 4bc, . A, = 2------ cos — tan — =------ sin —, 2, b+c, 2, 2 b+c, , h/2, , C, , h, , Since, AD ± EF and DE = DF and AD is bisector., => AAEF is isosceles., Hence, (a), (b), (c), (d) are correct answers., , T, , A, , 2h, AB, Now, in AABP, tan(a + P) = — =, AP, AC =2_, Also, in AACP, tana = —, AP ~ 2h, , 206. Let AB = AC = a and Z A = 120°., 2, , A Area of triangle = - a sin 120°, 2, where, a = AD + BD, = 73 tan 30° + 73 cot 15°, , ■+, , h, 2h, , 1, 2, , 1., , 4, , Now, tanp = tan [(a + P) - a], , 1, 1_1, _ tan(a + P) - tana _ 2 4 = 4=2, 1 + tan(a + p) tana 1 + -X- 9 9, 2 4 8, , =1+——, tan (45°-15°), x A x, , 2, , A, „ . A, 2 sin — cos —, 2, 2, B-C5!, cos, 2, . A, sm —, 2, (B-C, sin, , a, , => AE is HM of b and c., , 60°, , B-C, , cos, , I 2 J, , h, 208. Let AB = h, then AD = 2h and AC = BC = 2, Again, let Z.CPA = a, , AE = AD sec - =, 2 b+c, , 1, , fB + C), , i0°, , D,, , 209. Let AB = x, , a, , &, , a, , i, , c, , q, , D, , a, , 9, , O, , 1§>, M5°, , P, , P, , C, , B, , A, , k tan 45°-tan 30°, / r-, , — 1 + 73, , 7t-(0+a), , '1 + tan 45° tan 30°, , a = 1 + 73, , 313, , -x-q-, , \, , L73-1J, , M, -x, , q, , I, B, , In ADAM, tan(rc-0-a) = ——, x-q, , = 4 + 273, , :. Area of a triangle = - (4 + 273)2, 2, , a., , =>, , 2,, , = (12 4-7-73) sq units, , =>, =>, , tan(0 + a) = ——, q-x, q-x = pcot(0 + a), x = q-pcot(0 + a), , www.jeebooks.in, 207. Let a, b, c are the sides of AABC., , Now,, , b + c _k(sinB + sinC), k sinA, a, , ( cot0cota -1, cota + cot0, , q, , v cot a = —, P.

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www.jeebooks.in, 314, , Textbook of Trigonometry, , (icotO-ll, P, = <7~P, — + COt0, , Ip, , 211. We know that, ~^-=2R, sin C, , qcotB-p, q + pcot0, , J, , and, , [ qcos0-psin0 1, = <7"P ^gsin0 + pcos0 J, , 4, , Alternate Solution, Applying sine rule in zlABD,, q, , [vC=90°], , (i), , s -c, , r, s—c, , r = s-c, a+b+c, =>, r --------------- c, 2, => a + b - c = 2r, On adding Eqs. (i) and (ii), we get, 2(r + R) = a + b, , (p2 + g2)sin0, pcos0 + gsin0, , D, , 2, , 7t, => tan — =, , q2 sin0 + pq cos0 - pq cos0 + p2 sin0, pcos0 + qsin0, , AB, , c = 2R, C, r, tan — =------, , c, , -(ii), , 212. In ABAD, cos (90° - B) = —, , 0, P, , 4, , c f/&, , n-(0+a), B, B, , Vp2 + g2, , AB _, sin0, , sin {n -(0 + a)}, , AB =ylp2 + q2, , sin0, , sin(0 + a), ^/p2 + q2sin0, , AB =, , sin0 cosa + cos0sina, , (p2 + q2)sin0, and sina =, pcos0 + qsin0, , 210., , q, , cosa =, , P, , TAm7, , 90°-CH, D, , C, a, , =>, AD = c sin B, Similarly,, BE = a sin C, and, CF = b sin A, Since, AD, BE and CF are in HP., So, csin B, a sin C and b sin A are in HP., 1, 1, ,, 1, and, sin C sin B' sin A sin C, sin B sin A, are in AP., Hence, sin A, sin B and sin C are in AP., Alternate Method, ar (AABC) = | x BC x AD, =>, , =>, , By formula of regular polygon,, a, .it, , a, n, —7= sin— and — = tan—, 2R, 2r, n, n, , r, it, — = cos—, R, n, r 1, —=, R 2, r, n = 4 gives — =, R y/2, , n = 3 gives, , J_, , r %/3, n = 6 gives — = —, R, 2, , 1, A = -Xax, AD, 2, 2A, AD = —, a, , Similarly,, , BE= —, b, 2A, and, CF =—, c, Since, AD, BE and CF are in HP., 11, J, ., So, —, - and - are in HP., , a b, c, Hence, a, b and c are in AP., sin A, sin B and sin C are in AP., , www.jeebooks.in

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www.jeebooks.in, CHAPTER, , 1, , i, , Inverse Trigonometric, Functions, Learning Part, Session 1, • Inverse Trigonometric Functions, • Inverse Function, • Domain and Range of Inverse Trigonometric Functions, Session 2, • Property I of Inverse Trigonometric Functions, Session 3, • Property II of Inverse Trigonometric Functions, Session 4, • Property III, IV and V of Inverse Trigonometric Functions, Session 5, • Property VI, VII and VIII of Inverse Trigonometric Functions, Session 6, • Property IX of Inverse Trigonometric Functions, Session 7, ‘ • Property X, XI, XII and XIII of Inverse Trigonometric Functions, , Practice Part, • JEE Type Examples, • Chapter Exercises, *, , www.jeebooks.in, Arihant on Your Mobile!, Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.

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www.jeebooks.in, Session 1, Inverse Trigonometric Functions, Inverse Function,, Domain and Range of Inverse Trigonometric Functions, Inverse Trigonometric, Functions, Consider;, , .71 1 . 57t 1 . 137t 1, sin — = -, sin — = sm---- - = -,, 6 2 6, 2, 6, 2, , . 1771 *1 . ( llTt^l, 1 . ( lit, 1171, sin----- =-, sinl-------- =-, sin -—, =-,... etc., <, 2‘, 6', 6 7 2, 6' J 2, , Now, if the question is “which is that angle or real, , 1, 1, or arc sin, 2, 2, , 71. ., — =sin, 6, , {read as sine inverse, , It must therefore, be noted that sin-1 x is an angle and, denotes the smallest numerical angle, whose sine is x., , Similarly, cos-1 x and tan-1 x , denotes an angle or real, number; ‘whose cosine is x’ and ‘whose tangent is x’,, respectively, provided that the answer given are, numerically smallest available., , number whose sine is —?”, 2, , Note, , Then there are infinite answers i.e. there is no unique, answer., , 6, 2, 6, (ii) If there are two angles, one positive and the other negative, having same numerical value. Then, we shall take the positive, value., , siny = x, then for the given value of x = —, we have, 2, 771, , 6, , 6, , e.g., and, , 1171 71 571 1371, 6, , sin—= 1 but — *sin-1(1/2), , (i), , Thus, the problem is, if the trigonometrical equation is, , 6, , it, , 1, , cos—=—=, 4 V2, cos^-^=, , But we write cos', , i.e., , i, , 2_, , 12, , ^Hnotcos’U)=, , it, , 4, , for given value of x, we get infinite value of y., , Correspondence from the set, , Inverse Function, , {x | x 6 R; -1 < x < 1} to the set, {y | y G R; sin y = x} is one to many correspondence, , It cannot be a function., {'.* one-many and many-many are not function.}, However, if the question is “which is the numerically, , smallest angle or real number whose sine is -?”, 2, , Definition If a function, say f is one to one and onto, from A to B, then function g which associates each, element y E B to one and only one element x G A, such that, y=f (*)» then g is called the inverse function of f,, denoted by x=g(y)., , Usually we denote g=/’1, , 71, , (Read as/inverse), , x=/’1(y), , Then the answer is —., 6, , 71, , This is one and only one answer i.e. — is the unique, answer., 7t 1, In this case, the relation sin—=- is also written as;, 6, , 2, , Remark, If y=f(x) and x=g(y) are two functions such that/{g(y)}=y and, g{ftx)}=x then f and g are said to be inverse functions of each, other. Before we start with the definitions ofsin-1 x,cos‘1x,, tan-' x, etc., let us study the following discussion which will help, us to define these inverse functions., , www.jeebooks.in

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 317, , Domain and Range of Inverse, Trigonometric Functions, , ■, , 1. If siny = x, then y =sin 1 x, under certain conditions., Y., , (1,n/2), I Principal, /value branch, / ofy=sin'1x, 0, , 0<y<7t, , These restrictions on the values of x and y make the, function cos y = x one-one and onto so that the, inverse function exists i.e. y =cos-1 x is meaningful., Thus,, Domain : x G [-1,1 ], Range:, yG[0,7t], , 3. If tany = x then y = tan-1 x, under certain conditions., Y, , --------, , y= n/2, y~ tan-1/, , (—1, —rt/2), , -l<siny <1; butsiny = x., -1<x<1, Also,, siny = -l => y=-n/2, and, siny = l => y=n/2, Keeping in mind numerically smallest angles or real, numbers, we have -7t/2<y<7t/2, These restrictions on the values of x and y make the, function siny = x, one-one and onto so that the, inverse function exists, i.e. y =sin-1 x is meaningful., Thus,, Domain : x G [-1,1], Range: y G [-71/2,71/2], , y = -x/2, , tanyGR, xeR, Also,, -«><tany<oo, -7t/2<y <7t/2, These restrictions on x and y make the function,, tany = x one-one and onto so that the inverse, function exists, i.e. y = tan-1 x is meaningful., , Thus, Domain : xg R, Range :y G(-7t/2,7t/2), 4. Ifcoty = x, theny =cot-1 x, under certain conditions., , tv, , y=K, , (0. xJ2), , Note, (i) We can restrict the values of y in any of the intervals, [-3n/2, - k / 2], [-% / 2, n / 2], [ff / 2, 3n/2] etc., Corresponding to each such interval, we get a branch of, the function y = sin-1 x. The branch with range [—it/2. n/2], is called the principal value branch, whereas other, intervals as range give different branches of sin-1., (ii) The numerically least angle is called the principal value, of the function., , 2. Let cos y = x then y =cos, conditions, , Also,, and, , —X, , O, , x, under certain, , y= cot- x, =^X, , O, , cotyGR, xgR;, Also, -oo<coty <“=^0<y<7t, These restrictions on x and y make the function,, coty = x one-one and onto so that the inverse, function exists, i.e. y =cot-1 x is meaningful., , Thus, Domain: xG R, Range :y g(0,tc), , -1<x<1, cosy=-l=>y=7t, cosy = 1 =>y =0, , x, where |x| > 1 and 0 <y < n,, , 5. Ifsecy = x, theny=sec, y*n/2, , y, , Y., , (-1.K), , y = iJ2, y = cos-1/, , www.jeebooks.in, 0, , (1.0), , 0, , (1.0), y = sec-1/, , Here, Domain : x G R - (-1,1), Range : y G [0, it] - {ti / 2}, , X

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www.jeebooks.in, 318, , Textbook of Trigonometry, , !, , (ii) 3rd quadrant is not used in inverse trigonometric functions., , 6. If cosecy = x then y = cosec Tx,, , (iii) 4th quadrant is used in the clockwise direction i.e. - — < y <0., , |x|>l and-7t/2<y <n 12, y ^0, , where, , 2, , (iv) No inverse function is periodic., (v) If no branch of an inverse trigonometric function is mentioned,, then it means the principal value branch of the function., , Y, (1.JV2), , ■X, , o, , H.-n), , y = cosec-1x, , Here, Domain : x e J?-(-1,1), , Sol. Let y = tan cos, , Principal Values and Domains of Inverse, Trigonometric/Circular Functions, , Domain, , (i), , y = sin-1x, , -1<x<1, , (ii), , y=cos x, , -l<x<1, , (iii) y=tan x, , y= sec x, , (vi) y=C0t-1x, , + tan, , = tan, , —+, , It, , 7t, , 3, , 6, , = tan, , 1, , A, 71, , 1, , 6, , 3, , Range (Principal Value Branch), 71, , 71, , 0< y<n, , xeR, , (iv) y = cosec-lx x < -1, orx > 1, (v), , 1, , 2, , Range : y e [-7t / 2,7t / 2] - {0}, , Function, , I, , I Example 1 Find the value of, 1, ' 1_ ', tan cos, + tan, /3, 2, , I Example 2 Find the value of cos cos, , 71, , n, , n, , 6, , 7t, , +—, , Sol. cos cos, , n, 2, , 2 7, , 71, H—, , < 2 ,, , 6, , (57t 7t|, . ., = cost--- + ~f = cos(7t) = -l, I 6, 6I, , 2, , I Example 3 Find domain of sin“1(2x2 -1), , x<-l, orx £ 1, , 71, 0<y<7t; y*-, , Sol. Let y = sin-1(2x2 - 1), , xeR, , 0< yen, , For y to be defined -1 < (2x2 -1) < 1, , => 0<2x2 <2 => 0<x2 <1, Note, , => xe[-l,l], , (i) 1st quadrant is common to the range of all the inverse, trigonometric functions., , Exercise for Session 1, n irn T«ri——crw* «»r, , w~i—a<nn utriii, , ,» sr ti« v.:k, , Find the value of the following, 7. sin[f-sin-’Q)], , 2. cosec [sec“1(- 72) + cot~1(-1)], Find the domain of the following, , 3. y = sec-1 (x2 + 3x +1), , 4. y = cos, , /, I, , X, , 2, , \, , V+ x2 y, , www.jeebooks.in, 5. y = tan-1 (-Jx2 -1)

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , 5 71, From the graph, we can see that if 2tc < x < —, then, , Also, sin( 27t - 5)=- sin 5=sin(- 5), sin-1(sin(-5))=sin-1(sin(2n-5))= 2tc-5, , y = sin 1 (sin x) can be written as, y - x -2n, sin-1 (sin 7) = 7 - 2n, Alternate Method, We know that, sin-I(sin0)=0, if-7i/2<0 <7t/2, , (iii) We know that, cos-1(cos0)=0, ifO<0<7t., , Here, 0 = 10 radians which does not lie between 0 and, 7t. However, (471-10) lies between 0 and 7t such that, cos(47t-10)=cosl0, cos-1(cos10)=cos-1(cos(47C-10))=4tt-10, , Here,0=7 radians which does not lie between-n/2 and, 71/2., 71, , 321, , (iv) We know that,, , tan-1(tan0)=0, if—7t/2<0<7t/2., , 71, , But, 2tc -7 and 7 - 2tc both lie between — and —., 2, 2, Also,sin(7-2Tt)=sin(-(27t-7)) = -sin(27t-7) = sin 7, sin-1(sin7)= sin-I(sin (7-27t)) = 7 — 2tt, , Here, 0=-6radians which does not lie between -n/2, and n/2. But we find that 2tt-6 lies between -7t/2and, n/2 such that, tan (27t-6) =-tan6 = tan(-6), , (ii) Here, 0 =-5 radians. Clearly, it does not lie between, 7t »7t _, ,, rt, , rt, — and —. But 271-5 lie between — and —., 2, 2, 2, 2, , tan-1(tan(-6)) = tan-1(tan(2n-6)) = 2n - 6, , Exercise for Session 2, 1. Find the value of cos, 2. Write the value of sin, , (, , I, , cos — + sin, , 3J, , . 271A, sin — ., , 3J, , ( •, , I, , 5J, 3371^, , 3. Find the value of sin 1I cos----- ., , 5 J, , 4. Find the value of cos"1(cos 13)., 5tt , 3tt I., 5. Find sin-1 (sin0),cos‘1(cos0),tan’1(tan0),cot"1(cot0)for0el —, , 2, , Session 3, Property II of Inverse Trigonometric Functions, Property II, , Proof We know that, if f: A —> B is a bijection, then, /-1 :B—> A exists such thatyb/-1(y) = f (f~X(y))=y for all, , (i) sin (sin-1 x) = x,, , for all x g[-1,1], , (ii) cos (cos-1 x) = x, , for all x 6 [-1,1], , (iii) tan (tan-1 x) = x, , for all xG R, , (iv) cosec (cosec-1 x) = x, , for all x G, , (v) sec (sec-1 x) = x,, , for all x G (—°°, — 1] O [ 1, °°), , Clearly, all these results are direct consequences of this, property., l]u [1,°°), , Aliter Let 0 G [- it / 2, it / 2] and x G [-1,1] such that, sin0 = x., Then, 0=sin-1 x, , .'. x=sin0 = sin (sin-1 x), , www.jeebooks.in, (vi) cot (cot-1 x) = x,, , for all x G R, , Hence, sin (sin-1 x) = x for all x e[-l»l], , Similarly, we can prove other results.

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www.jeebooks.in, 322, , Textbook of Trigonometry, , y, , Graphically they can be Shown As, , J k, , (i) y = sin (sin-1 x) =cos(cos-1 x) = x,, , x G [-L1], y G [-1,11 y is aperiodic, -!•, , y, , 1, , I, , 0, , 1, , >x, , -1, , 1, , I, I, , >x, , +1, , 0, , i, , 371, , 1 3 Jv, , I Example 6. Find the value of cosec cot cot —, 4, , Sol. Let y = cosec|cot^cot *—)•, (ii) y = tan (tan 1x)=cot(cot ’x) = x, x 6 R,y G R-, y is, aperiodic, , v, , cot(cot-1 x) = X, V X 6 R, , ( -i 3n, cot cot —, k, 4, , y, , •(>), , 371, 4, , y 371^ r~, Now, from Eq. (i), we get y = cosec — I = V2, I 4), , •x, , o, , I Example 7. Prove that, sec2 (tan-12)+cosec2 (cot-1 3) = 15, Sol. sec2(tan-12) + cosec2(cot-’3), , (iii) y = cosec (cosec Jx) = sec (sec 1 x) = x,, , |x| > X|y | > 1; y is aperiodic, , = {1 + tan2 (tan-12)} + {1 + cot2(cot-13)}, = 1 + (tan (tan-’2))2 +1+ (cot(cot-13))2, , = 2 + 22+32 = 15, , Exercise for Session 3, Evaluate the following:, 1. cos sin sin-1— •., , k, , 6J, , 3n, 4, , 2., , sin cos cos-1, , 3., , sin2 cos, , 4., , tan2(sec-12)+cot2(cosec-13)., , 5., , Find the solutions of the equation cos(cos-1 x) = cosec (cosec-1x)., , 1, 21, + COS I sin, 2, , 1, 3, , www.jeebooks.in

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www.jeebooks.in, Session 4, Property III, IV and V of Inverse Trigonometric Functions, Property III, , y = 7t - cos-’(sin5), , (i) sin-1 (-x) = -sin-1 (x), for all x G [-1,1 ], , cos---- 5, , = 71 - COS, , (ii) cos~’(—x) = 7t -cos-1(x), for all x g[-1,1], , (iii) tan-1 (- x) = - tan-1 x , for all x G R, , Here, -2it<—-5<-lt, , (iv) cosec-1(-x)=-cosec-1x, for all xg (-«»,-l]u[l,°°), , and the graph of cos-’(cos x) is as, , xg, , J, , 12, , y, , (v) sec-1(-x)=7t-sec-1 x, for all xe(-“-l]u[l,°°), , (vi) cot-1(-x)=7r-cot-1 x for all, , cos-1 (cos x), , R, , ?<■................... K, , Proof (i) Clearly, -x G [-1,1] for all x G[-1,1 ], Let, , Then,, , sin ’(—x)=0, , k!2, , \V, , y I, , -.(i), , —y-—i----- i-2n 3k, 2, , — x =sin0, , ...(i), , 12, , z, , +, K, , 4K, , 2, , 2, , Z, , to, , 7, , 4K, , -4—, 3k, , 2k, , 4------ H, 5k 3k, , =>, , x = -sin0, , =», , x=sin(-0), , From the graph, we can see that if - 21t < x < - 7t,, , =>, , -0 = sin-1 x, , then y = cos-’(cos x) can be written as y = x + 2n., , {•/xe[-l,l]and-0G[-n/2,rt/2], for all 0G [-71/2,71/2], 0 = -sin, , x, , Now, from Eq. (i), we get, tt, f57t _,, c 3-y = 7t ------- 5 => y = 5------I2 J, 2, , I Example 9. Evaluate the following, , cos-1 (- x) =0, , =>, , x = -cos0, , =>, , cos(i -5)H, , ../ii), , (ii) Clearly, - x G [ -1, l]for all x g[- 1,1], , - x =cos0, , tt -5, |-5| + 27t = |5—, 2, , cos, , sin-1 (- x)=-sin"’(x), , Then,, , (, , (i) sin, , ■, , (, , -3te, Sin | —, , (ii) cot-1 (cot(-4)), , 4, , So/, (i) sin, , x = cos (tc - 0), , 7, -371^, sin|---- = sin, 4 7, , cos-1 x = 71 -0, {•/ x g [ -1,1 ] and 7i - 0 e [ 0,7t ] for all 0 g [0,7t ]}, , . 371, sin —, 4, , -sin, ■, , = -sin, 0 =7t - COS-1 X, , 2, , /.From the graph, , From Eqs. (i) and (ii), we get, , Let, , 2, , (, , 71, , 71, , sin| it—, 4, , =-sin, , . 7t, sin —, 4, , 7t, 4, , (ii) cot-’(cot(-4)) = cot"’(-cot4) = 71-cot ’(cot4), , ...(ii), , = It—C0t-’(C0t(7l4-(4“7t))), , From Eqs. (i) and (ii), we get, , = n-cot”’(cot(4-n))=2n-4, , COS~’(—x) =71 —COS-’ X, , Similarly, we can prove other results., , I Example 10. Evaluate the following, , I Example 8. Find the value of cos-1 {sin (-5)}, , (i) sii, , 'n . ., —sin, , J, , 2, , www.jeebooks.in, Sol. Let y = cos-’ {sin(- 5)}, , = cos"’(- sin 5), , cos (- x) = 7t - cos-1 x,|x| < 1, , /, .... .71, (n)sin ——sin, , 2 >)

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www.jeebooks.in, 324, , Textbook of Trigonometry, , ( 71, , 1, , Sol. (i) sin---- sin, , 71, , ., , . ., , = sin — +sin, , V, , I2, , . / 71, , From Eqs. (i) and (ii), we get, -if C, -i, cos, =sec U), , 1, 2 7, , 71 1, , /, /, 73, (ii)sin----sin, 2, \2, , X, , 73, , .271, 271, , =sin —+ — | =sin — =, <2 6, 3, 2, 7 71, r /, = sin — + sin, 2, , -T, , (iii) Let cot, , x =0. Clearly, x e R, x *0 and 0 6 (0,7t), , Now, two cases arises, , Case I When x >0, In this case, 0 e (0,7i / 2), , .11 It 1, .571, . 5n 1, =sinl —+ — = sin—=, 12 3, 6 2, , cot-1 x=0, X=COt0, , Property IV, (i) sin, , (ii) cos, , — = tan0, x, , | — | =cosec-1x, for all xe(-<»,- Ijufl,®0), <x, , fl, , (iii) tan, , =sec, , -(ii), , x, , x, for all xg(-“- 1]o[1,°q), , 1, , cot, , x, , - 71 + COt, , {-.•06(0,71/2)}, , From Eqs. (i) and (ii), we get, , x.for x >0, , 1, , tan, , x,for x<0, , =cot, , x, for all x >0., , X, , Case II When x <0, , x=0, , (i) Let cosec, , ...(i), , x =cosec 0, 1, , x=cot0<O}, , In this case, 0e(7i/2,7t), 71, , Now,, , n, , — <0<7l, 2, , - n, , — =sm0, x, , 2, , •.•xe(-oo,-l]o[l,~) => -6[-l,l]-{0}, , <0 — 7C <0, , 0 -Tie (-71/2,0), , X, , cot-1 x=0, , cosec-1x=0 => 0e[~7t/2,7t/2]-{0}, , X=COt0, , 0=sin, , x, , — = tan0, x, , From Eqs. (i) and (ii), we get, 1^1, =cosec xx, sin, x, (ii) Let sec, , 1, , 9 = tan, , Proof, , Then,, , ■(0, , tan(7i-0) = -tan9}, , —=-tan(7t -0), x, , x=0, , — = tan(0-7t), x, , ...(i), , Then, xe(-°°- 1]u[L®°) and0e[O,7i]-{7t/2}, 0—7t = tan, , Now, sec-1 x=0, , 1, x=sec0 => — =cos0, , tan, , X, , 0=cos, , {•.•0-7tG (-71/2,0)}, , x), , 1, , •••(ii), , x, , — |=-7t+0, XJ, , •..(Hi), , From Eqs. (i) and (iii), we get, , n = -7t +cot, , tan, , VX =(-°°,-l]u[l,oo), , x, if x <0, , xj, , www.jeebooks.in, 71, , =>—e [—1,1] —{0} and0e[O,7i]-- — 2, , Hence,, , tan, , x,for x>0, , 1, , cot, , x, , -71 +COt, , x.for x<0

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www.jeebooks.in, i, , Chap 04 Inverse Trigonometric Functions 325, , I Example 11. Evaluate the following, (i) sinf cosec'15, (ii) cotf tan'1 I, 3, I, 4, /, , 5, , Sol. (i) sin cosec, , = sin sin, , 3, (ii) cot tan, , 3, , 4, , 3, , 3, , 5, , 5, , .!, , .-i 4 i, , 4, , I, , 37, , 3, , - =, , = cotl cot, , =>, , 6+cos, , x=7t/2, , From Eqs. (i) and (ii), we get, sin, (ii) Let, tan, Then,, , 2, , x =0, , ..<i), , —I, , Sol. Let y = tan <! cot, , n ~ n, — <0<—, , I 3, , f—-0 G(0,7t), k2, , 2\, , 3J, , Now, tan, , 2, 3, , /, , x=0, x =tan0, , x=cot(7t/2-0), , cot x = tan, , — ifx>0, x, y = - tan tan 3 I, , 3, 2, , 7t / 2 — 0e (0,7t)}, , x=--0, 2, , cot, , -2ry=-, , 7t, , ,(ii), , x=—, 2, , 0+cot, , From Eqs. (i) and (ii), we get, , Property V, , tan-1 x+cot-1 x=7t/2, , (i) sin-1 x+cos, , x = it / 2, for all x g [-1,1], , (ii) tan-1 x +cot, , x = 7t / 2, for all x e R, , (iii) sec, , 71, , 0<— —0<7t, 2, , Eq. (i) can be written as, , y = - tan cot, , M, , 7C, , cot-1(- x) = 7t - cot-1 x, x e R, , y = tan 171-cot, , 2, , — <-0< —, 2, 2, , 7,, , —(i), , 3 )., , {vxeR), , &E(-n/2,n/2), , 2, , I Example 12. Find the value of tan • cot, , 71, X= —, , X +COS, , (iii) Let, sec, , Then,, , 7t, , x +cosec, , Ui), , x=0, , 0 G [0,71 ] - {7t / 2} {•/ X G (- oo, - 1] U [ L °°)}, , 0 <0 <71, 0 *7t/2, , x=—, for all xg(-©o,-1]u [L°°), 2, , -7t<-0 <0,0*71/2, , Proof, ■ (i) Let,, , sin, , x=e, , Then,, , 0 G [-7t/2,7t/2], , =>, , - n/2 <0 <71 /2, , =>, , -7T/2<-0<7t/2, , =>, , Now,, , x=0 => x=sin0, , x=cos---- 0J, , U, , cos, , 2, , ), , 71 -0n, x=—, 2, , 2, , 2 2, , \ r -71, , {VXG[-1,1]}, , O<--0<71 => _-0G[O,7t], 2, 2, sin, , 2E_0;to, , ...(i), , 7t, , --0 G -- > > --0*0, 2 2 2, , U } L Now,, , sec, , x=0, , x=sec0, x=cosec (ti/2-0), , cosec, , x =7t/2-0, 71 71, . TIE, I \2 J L -2*2, , 0+cosec~1x=7t/2, , --0*02, , www.jeebooks.in, {vxg[-1,1] and (7t/2 — 0) G [0,7t]}, , From Eqs. (i) and (ii), we get, , sec -1 x+cosec-1x=7t/2, , ...(ii)

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www.jeebooks.in, 326, , Textbook of Trigonometry, , I Example 13. Find the value of of, , cos cos, , sin (2 cos-1 x + sin-1 x) when x = -, , /, , 1, 5, , -11, , 5, , Now, from Eq. (i), we get y = |, , Sol. Let y = sin [2cos-1 x + sin-1 x], x + cos, , sin, , y = sin 2cos, ., , *, , I Example 14. Solve sin x-cos x = cos"1, , ■1x = p|x|<l, 7t, , X +----- cos, 2, , = sin — + cos, , -i, , 2, , _i, , X, , So/. We have, sin-1 x-cos-1 x = cos, , x = cos(cos 1 x), , sin, , i, , x-cos, , Also, sin 1 x + cos, , x=5, , i, , 73, 2, , 73, 2, , n, x=—, , •(i), , 6, , 7t, X=—, , ■(ii), , 2, , y = cos cos, , 7t, , -ill, -, , On solving Eqs. (i) and (ii), we get sin x = —, 3, , ...(i), , 5J, , and, , cos(cos-1 x) = x if x e [-- 1,1], , cos, , it, x=—, 6, 5/3, , and, , x = — is the only solution., , Exercise for Session 4, Evaluate the following, , 1. tan-1 !tan f-—, , I, , I 8 Jj, , 2., , 3., , sec cos, , 4. cosec sin, , I, , - -7=, , I &)), , ., , 5., , cos, , 6., , If sin' x = —, for some x e (-1,1), then find cos, 5, , x., , 7., , If sec 1 x = cosec 1y, then find the value of cos, , —+ COS, , 1, , 2, y, , 8., , 1 = n/2,if x >0, Prove that tan-1 x + tan-1 —, -n/2,if x<0‘, x, , Solve the following, 9. 5tan-1 x + 3cot-1 x = 2it, , www.jeebooks.in, 10., , 4sin-1 x = n -cos-1 x

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www.jeebooks.in, Session 5, I, , i, , Property VI, VII and VIII of Inverse Trigonometric Functions, Property VI, , tan" x + tan, , ' x+y ', , y = tan, , (i) tan-1 x + tan-1 y, , tan, , j-*y,, Case //When x <0, y <0 and xy <1, , (£±Z ,if, , In this case;, , 1-xyJ, /, , it + tan, , x<0, y<0and xy<l, , x+y, , ,if x>0,y>0 and xy >1, J-xy), x+y, , if x < 0, y < 0 and xy > 1, ^1-xy, , —it + tan, , i±i<0, , =», =>, , 1-xy, , {from Eq. (i)}, , tan(A + B)<0, , => A + B lies in II quadrant or in IV quadrant., (ii) tan-1 x-tan-1 y, , tan, , =} A + B lies in IV quadrant., vx<0=>-7t/2<A<0, , x-y ,if xy >-l, ,1 + xy,, x-y Lif x>Q,y <0and xy <-l, , It + tan, , — <A + B<0, 2, , 1+xy J, , tan(A + B)=i±i, , ,if x <0,y >0and xy <-l, , -7t + tan, Proof, , A + B = tan, , (i) Let tan-1 x -A and tan-1 y = B. Then,, x = tan Aandy = tanB and A, Be(-it/2,7t /2), tanA + tanB x+y, tan (A + B)=----------------- =----- —, 1 - tan A tan B 1-xy, , tan, , x + tan- y = tan, , Case III When x > 0 and y < 0 or x < 0 and y > 0, x > 0 and y < 0 =* A G (0, it 12), , tan(A + B)>0, , => A + B lies in I quadrant or in ID quadrant., , 0<A + B<7t/2, , and, , B G (- 7t / 2,0), A + B G (- 7t / 2,7t / 2), tan(A + B)=^^1-xy, /, \, x+y, A + B = tan, , {from Eq. (i)}, , 4 „, • =>0<A + B<7t, , y >0=>0<B<7t/2, , tan(A + B)=^, 1-xy, / x+y, . \, A + B = tan, , / x +y, . \, , Consider,, , x+y n, x>0,y>0andxy<l => ——>Q, l-xy, , •.•x>0=>0<A<7t/2], , J-xyJ, J“*y,, , Case /When x>0, y>0 and xy <1, , =>, , ' x+y, , ...(i), , Now, the following cases arises., , =>, , {from Eq. (i)}, , 1-xy, , i.l + xyy, , In this case,, , =>-7C<A + B<0, , y <0 => —71/2 <B <0, , < x +y>, tan-1 x + tan"1 y = tan-1, , J-*y,, , {from Eq. (i)}, , Similarly, if x <0 and y >0, we have, , www.jeebooks.in, f, , tan, , x + tan, , y = tan, , x+y, , U-*yJ’

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www.jeebooks.in, 328, , Textbook of Trigonometry, , It follows from above three cases that, tan, , x + tan, , y = tan, , => tan(A + B) lies either in I quadrant or HI quadrant., , x +y, , 1-xy J, , if xy <1, , {■/x<0, y<0=> A, BG(-7t/2,0)=> A + Be(-7t,0)}, ==> A + B lies in III quadrant., , Case IV x >0, y >0 and xy > 1, , =>, , In this case, we have, , =>, , -7t<A + B<-7i/2, , 71-71 <71+(A + B) <71 -71/2, 0 <7t +(A + B) <71/2, , =>, , x>0, y >0 and xy >1, , tan(A + B)=iii, , Now,, =>, , {from Eq. (i)}, , 1-xy, , 1-xy, , tan (tc + (A + B)) = X, 1-xy, , tan(A + B)<0, <fromEq. (i), tan(A + B) = X +~^ ■, 1-xy, , / x+y, , 7t + A+B = tan, , J-xy, , => A + B lies in II quadrant or in IV quadrant., , => A + B lies in II quadrant., , ^x+y"], <i-*yj, , A + B = - it + tan, , vx>0,y>0=>A,Be(0,7t/2), , =>A + Bg(0,ti), , x + tan, , tan, , x+y', , y = - 7t + tan, , l-xy;, , 7t/2<A + B<7t, 7t, , (ii) This property can be proved by replacing y by -y in, above results., , — -71 <(A + B)-7t<0, 2, -7t/2<(A + B)-7i<0, , Remark, , /. tan(A + B)=^^1-xy, =>, , If xvx2, x3.... xn eR, then, tan-1 x3 + tan"1 x2 +...+ tan"1 xn = tan, , - tan{7t-(A + B)} = -y-—■, 1-xy, , where, denotes the sum of the product of xv x2.... xn taking A, at a time., , {•/ tan(7i-(A + B))=-tan(A + B)}, , =>, , tan((A + B)-tt) = -^——, 1-xy, , A + B-7i = tan, , A + B=7t + tan, , x + tan, , tan, , I Example 15. Prove that, , (i) tan-1-+tan-1 —=tan, 7, , x+y, , 13, , 1, , -jf x+y, ^i-xy,, , (iii) tan, tan -l-+tan-1i+tan_, 5, , 2, , x +y, , X, , -i 3, , ’, , -i 3, , (iv) tan- -+, -+tan, tan —tan', , x<0, y <0and xy>l, x+y, ----- — >0, 1-xy, , 1 _7l, , 8~i, 8 _7C, , 19 ~4, 1, (v) tan-1 i+tan-1-+tan- -+tan, 5, 7, 3, 4, , 5, , In this case, we have, , =>, , 9, , 3tt, , i, , 4, , Case V When x<0, y <0 and xy >1, , =>, , 2, , (ii) tan 12+tan 3=—, , l-xy;, , y=7t +tan, , 5i S3 + S5..., J—52 +, , Sol. (i) LHS = tan, , - + tan, 7, , tan(A + B)>0, , 1 _7t, 8~^, , [11], -+—, t, -i, 7 13, — = tan, 13, 1--X —, 7 13 J, , ■/tan x + tan y=tan, , x+y, ;if xy<A, 1-xy, , www.jeebooks.in, ■ from Eq. (i), tan (A + B) = -y+^ >, 1-xy, , = tan, , 20, , 90, , = tan, , - |=RHS., 9J

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , I Example 16. Prove that, , 24-3, , (ii) tan-12-Ftan-13= n + tan, , x+y, , 22,,2_, 2, r2, , 2, , x 4-y 4-2 =r, , tan-1Xj4-tan-1x2-Ftan-1Xj4-... -Ftan, , = 7t-7t 14 =371/4, , 1, , (iii) tan-1-4-tan-1-4-tan, 2, 5, , 1, , —-F-, , 2 5, , 1, , 4-tan, , 8, , 1--X2 5), , z, , using; tan 1x-Ftan-1y=tan, , x+y ', , 1-xy,, , = tan, , 1.9 J, , 651, . n, — = tan (1)=65), 65, 4, , 1, , 19, , ■13, , ■131, , f 3 3, , x, , _, , 19, , —F —, , = tan, , 5, , 3 3, 1 — x-, , 4, , k, , 7t, , = tan-(°°)=— = LHS, 2, , 8, , = tan -4- tan - -tan, 4, 5J, 4, , k, , where r2 =x2 +y2 +z2, , £, , 3, — tan, 5, , 3, , (iv) tan- -4-tan, 4, , , if xy<l ■, , 8, , ( 7 1 >, -4—, 9 8 = tan, = tan, . 7 1, 1—x9 &), , \, , yz | xz xy xyz, xr yr zr r3, = tan, x2+y2+z2^, 1r2, k, 7 7, k, f, r2, ! yz ,, xyz, l^zr xr yr r3 ), = tan, (x2 4-y2 +z2)-(x2 +y2 4-z2), , 1, , I - I 4-tan, , <S,-S3-f...>|, , where, St denotes the sum of the product of xt, x2,.... x,n, taking k at a time. Therefore,, / x, xz, LHS = tan, 4-tan, 4-tan, , 4-tan ’8, , 5, , 2, , = tan, , 8, , tan-1-4-tan-.11, , = tan, , ,, , Sol. We know that,, , = 714- tan-1(-l), , f 1, , It, , tan -4-tan —4-tan — = —, where, xr, yr, zr 2, , J-xy,, , =I, k, , xy, , i zx, , i yz, , 1-2x3, , using; tan-1x-Ftan-1y=7t-Ftan, , 329, , I Example 17. Show that (tan 114- tan 12+ tan 1 3) = it, Sol. tan-12+ tan-13=n + tan, , -tan, , ' 24-3 '', , {as (2) (3) >1}, , <l-2x3>, , = 714-tan-1(-l)= n-tan"’(l), , 5), , ...(i), , tan-1(l)4- tan-1(2)-F tan-1(3)=tan-I(l)-F 7t - tan"’( 1), , = tan, , 271 — -tan, 11J, , 4251, , = tan, , 425 J, , 8, , 11 19, = tan, 27 8, 19, 14- — X —, I 11 19j, , = tan, , (v) I tan-1-4-tan - |4-| tan -4-tan 1, 3, 8, I, 5, 7J, Z, ( 1 -1 >, 1 1, -4-5 7 4-tan, 3 8, tan, , k, =>, , 5 1), , tan, , 6, , — 4-tan, ,17j, , I Example 18. Solve for x;, , tan"1 (x 4-1)4-tan-1 x4-tan"1(x-1)=tan“1 3, , 4, , l-lxl, , {using Eq. (i)}, = 71, , l-lxl, , k 3 8), •if 11^1, , 123, , So/. tan-I(x 4-1) 4- tan-1 (x - 1) = tan-1 3- tan-,x, X4-14-X-1, , => tan, , l-(x4-l)(x-l)>, , = tan, , 3-x ', , , when x2 -1<1, , 14-3x>, , and 3x<l, =>, , 2x, , tan, , = tan, , 3-x ', , , when- 4i<x<42, , 14-3x>, , 1, x<-, , and, , www.jeebooks.in, Z, , tan, , 6---11, ---- F —, 17 23 = tan, . 6 11, 1----- X —, , 17 23 J, , 3251, _j, It, — =tan (1)=—, 325 J, 4, , =>, , 3, , 2x, , 2-x2, , 3-x ,, r, 1, ------ , when - >J2 < x < 14-3x----------------------3

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www.jeebooks.in, Session 6, Property IX of Inverse Trigonometric Functions, Property IX, , f, , -sec, , (i) ForO<x <1,, , sin, , x =cos, , = cot, , X, , 1, , Vl-x 2, 1-x2, , = tan, , Again,, , k 1-x2, sin0, n sin0, tan0 = ------=, , -Ji-sin2 0, , COS0, , tan 9 =, , = sec, , 2, , 2, , x, , \, , x, , P, , = cosec, , {'/ 0 = sin 1 x}, , x, , 0 = tan, , X), , k, , (ii) For 0 < x < 1,, f, , 2 =tan, , cos" x = sin, , k, X, X, , = cot, k, , = sec, , /1-x7, , f, , 1-x2, , X, , sin' x = tan, , x, , k, /, , 7, , P, , sin, , x = tan, , X, , (1, , = cot, , — cos, , l + x2 y, , = sec ^(Jl + x2), , 1, , sin, , -y 1 + x, k, , x, , 1, k, , 1+x2 ?, X, X, , = tan, k, f, , 2, , = cot, , X, , — forx>0}, x, , x =cot, , 1-x2 =sec, , x =cos, , \xj, , - cosec, , 7, , /, , f, , /, , X, , k, , x, , /, , k, , 1-1-x, --- FA, 2, , Hence,, , 7, , x, , = cot, , {•/ tan, , 2, , (iii) For x > 0, tan-1 x = sin-1 —, , (, , k, , X>, , k 1-x, , x, , x, , 1, , =cosec, , {•/ 0 = sin 1 x), , k, , 1^,, x, 7, , Proof, (i) Let, sin-1 x = 0. Then x = sin0, Now,, , cos Q = yjl~ sin2 0, cos0 = Jl - x 2, , =cosec, Similarly, the other results can be proved., , I Example 29. Evaluate the following, 3, (i) sin^tan"1, (ii) sin cos, , www.jeebooks.in, 0 =cos, , => sin, , x =cos, , 1-x2, , x2, , 5, , (iii) cosf tan-11, , (iv) sinfcot 1 x)

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www.jeebooks.in, 336, , Textbook of Trigonometry, , I, 3, So/, (i) Let y = sinl tan”1 —, , 3 - x/5 _ (3 - x/5)2, , —(i), , 3- x/5, , 4, , Note, tan — = ±, 2, x, , To find y we use sin (sin-1 x) = x, - 1, < x < 1, For this we convert tan-1 x in sin, , x, , 3/4, Jl+(3/4)2 ,, , sin, , 3^, , y = sin sin, , 5, , 3, , y=i, 2, , • -14, sm —, 5, , (ii) cos - = sin, 5, sin cos, , (iii) tan, , 14, 5, x, , = sin sin, , 1, , = cos, , 1X, , 3, , 4, 5, , 4, , 1, , = sin, , x, , I Example 31. Find the value of, i, 1, cos(2cos x + sin- x) when x = Sol. Let y = cos [2cos 1 x + sin-1 x], , 4, 5, , v, , 7, , -13, cos tan — = cos cos, 4, , (iv) Let, cot-1 x = tan, , x, , = cos, , 2, , 4, , 4, , 5, , 5, , 1/x, <y]l + l/x2, , = sin, , 1___, kVx2+l, , f1, I Example 30. Find the value of tan -COS, , I2, , Let cos, , y = -sin sin, , 1, ■Jl + x 2, , sin(cot-1 x) = sin sin, , I2, , 1, Sol. Let y = tan - cos, , 75, , /, , 1-, , 11, , x, , 7, x/24, , = -sin sin, , /p, , 5, , 5, , 3, , I Example 32. If sin 1 x+sin“1 y + sin 1z = n, prove, that x4 +y4 +z4 + 4x2y2z2 = 2(x2y2 + y2z2 + z2x2), , ...(i), , 3>, , Sol. We have, , --- = 0 => 0 £ | 0, —, 3, 2, , =>, , and cos 8 = —, 3, , .‘. Eq. (i) becomes, y = tan, , sin-1x + cos’x = — ,lx| < 1, 2, 7t, _!, y = cos 2 cos X +---- COS X, 2, 71, -1, = cos — + COS X = - sin (cos ’x), 2, 1, X=5, 1, y = - sin cos, 5, , c, , /, , /, , ...(iii), , 7, , 0, tan — > 0, 2, So, from Eq. (iii), we get, r3-_ 'T, tan—=, 2 x 2, Now, from Eq. (ii), we get, /, 3y=, , Now, from Eq. (i), we get, or, , 2, , 0, 6e fi 0, —, | => —e | 0, —, _ i, 4, V ' 2), 2, , X, , Here, tan, , 3-7T, , 0, , (ii), , 2, , sin-1 x + sin-1 y + sin-1 z = n, , =>, , sin-1 x + sin-1 y = 7t - sin-1 z, , =>, , cos(sin-1x + sin-1y) = cos(ti - sin-1z), cos(sin-1 x).cos(sin-1y) - sin(sin-1 x).sin(sin-1y), , = - cos(sin-Iz), , www.jeebooks.in, tan —, 2, , 1 - cost), 1 + COS0, , ____ 3_, 1+^, , 3, , 1-x2, , •71-/,2, , =, , z2, , {■■■ cos(sin-1 x) - cos(cos-17, , x 2}

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www.jeebooks.in, JEE Type Solved Examples:, Single Option Correct Type Questions, • Ex. 1. Let f(x) = sinx +cosx + tanx + arc, , • Ex. 3. If the equation 5 arc tan(x2 + x + k) + 3, , sin x + arc cos x + arc tan x. IfM and m are maximum and, minimum values of f(x), then their arithmetic mean is, equal to, , arc cot(x2 + x + k) = 2n, has two distinct solutions, then the, range ofk, is, , (a) —+ cost, 2, , \ 4, , (b) — + sinl, 2, , (c) —+ tanl + cost, 4, , (d) — +, + tanl, tan1 + sinl, sin 1, (d), 4, , As, tan" a + cot a=—Vaefl, 2, , Sol. (a) Domain of f is [-l,l];/(x)=sinx + cosx + tanx + sin"1 x, , + cos“’x + tan"’x, , >i, , and /tolmax, , Now,, , It, , For required condition, put D > 0, 1 -4(fc-1)>0 => 5-4Jt>0 => k<^, =>, , . ., , ., , — + n — = — + cosl -sinl - tanl, 2, 4 4, = /(I) =sinl + cosl + tanl, it n 3n, ., ... ., + —+ —= — + cosl + sinl + tanl, 2 4 4, M+m n, = — + cosl, 2, 2, , • Ex. 4. Iff(x) = x11 + x9 - x,77 + x3 +1 and, , /(sin-1(sin8)) =a, a is constant, then /(tan-1(tan8)) is, equal to, , (b)a-2, (d)2-a, , (a) a, (c)a + 2, So/.(d)f(x) + /(-x)=2, , x, , f 5, , • Ex. 2. The value of 5 • cot Z cot-1(/c2 + k +1) is equal, 7, to, , (a)j, , x2 + x + (fc-l) = 0, , =>, , Hence, f'(x) > 0 => f is increasing, => Range is [/(-I),/(I)], /(x)|rain=/(-l) = -sinl + cosl-tanl, 7t, , tan"’(x2 + x+fc)=^-=>x2 + x + k = l, , =>, , 1, 1 + x2, , 'u/zTT, , 7t, , Now,, , (sin-1(sin8))=37t -8=y (say), , and, , (tan"1 (tan8))=(8 - 3n)=-y, , Hence, Given, , f(y)+f(-y)=2, f(y) = a => f(-y)=2-a, , (b)7, , /, , ., , Sol. (b) Consider, , (fc + l)-fc, , zL tan, , 1 + ^+1), , k=l, , s, , = E tan-1(k +l)-tan“’fc, k=l, , 7j = tan-1(2) - tan"’(l);, , Now,, , 3, , X, , 7, 5 /, , 4, , T2 = tan"’(3) - tan”’(2) and so on, , I, , x4 - — + —, , 3 ’, , 7, , ), , = —, where 0<\x\< 41, then number of, 2, , Sol. (c) sin, , 7, , (d)4, , (c)3, , (a) 1 (b)2, r4, X, , x6, X, , 3, , 9, , x2---- + —., , 5, , + cos, , f 4 x8, I, 3, , X, , k=l, , 5, , +COS, , values of ‘x’ is equal to, , Hence, X cot-1(fc2 + k +1) = tan~’(6) - tan-1(l), = tan, , 9, , ., , X, , 12, , 8, , X, , 6, , x2 - — + —., , • Ex. 5. If sin, (c)-7, , ’ 4, , 3n, Sol. (b) We have 27t = — + 2 tan-1 (x2 + x + k), 2, , 71, , /'(x) = cosx-sinx + sec2x + 0 +, , 5, , » oo —, , (d), , (b) -oo,, , => x=y, , J., , x12, 9, , x, 7, , n, 2, , y, , x2, x4, 3x2, 3, =>, x2 = l+', 3 + x2 3 + x4, 1+—, 3, 3, 9 + 3x4 =9x2 + 3x4 => x2 = l, , www.jeebooks.in, = cot, , 7, =7, 5cot cot, 5, , =>, , 5>, , Thus,, x = 0,1 or -1, Hence, number of values is equal to 3.

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www.jeebooks.in, E►, , 346, , Textbook of Trigonometry, , • Ex. 6. Suppose3 sin ^logjxJ+cos \log2 y) =n/2 and, sin-1(log2x) + 2cos-1(log2y) = 1171/6 then the value of, , • Ex. 10. If the mapping f(x) = mx + c,m>$ maps [-1,1], onto[0,2], then tan tan, , x~2 + y~2 equals, (a) 6, , (b)7, , (c)5, , (d)L, 1, , J, 1-2, , (d)/fy, , Sol. (d) Clearly, f(x) = x + 1 (As, -1 <x< 1, 0 < x +1 <2) and, /(x) = mx + c, 1 tan -i ■ 1, Now, tan | tan-1 - + tan -+, 18, 8, I, 7, ( 1 1 \, , 1171, , 3a + b = — and a + 2b =---2, 6, a =---- and b = 7t, 6, , 7__ 8, , = tan tan, , 1, 1, 1, ,, 1, 11,, x = -n=andy = - =>— + — =6, Ji, 2, x2 y, y2, , Hence,, , 7, , (c)O, , Sol. (a) Letsin-1(log2x) = a and cos-1(log2y) = b, ., , -+ cot-18 + cot-118 | is equal to, , k, = tan tan', , • Ex. 7. Range of f(x} = sin-1 log[x] + log(sin-1[x]),, , !-lxl, 7, , 1, + tan -i —, 18, , 87, , 7, , fl5), , _1_, , 155 J, , 18, , — + tan, , where [] denotes GIF is, , (a) 1, , (b)2, , (c)0, , (d)llogy, , z, , 3, 1, —+ —, 11 18, = tan tan, .1-----3x —1, k, ii is;;, \, . _,1>, 1 7-2^, = tan tan- =- = j —, k, 3J, 3 I 3 J, , Sol. (d) Domain of f(x) is [1, 2)., •, , B, , (1, , .*. Range is < log — k, I, 21, s, • Ex. 8. Z sin-1(sin(2n -1)) is, , (, , • Ex.11. If(sin 1a)2 +(cos 16)2 +(sec ’c)2, , n=1, , 5ti 2, + (cosec-1c/)2 =----- , then the value of, , (b)2, (d)4, , (a)1, (c)3, , 2, (sin-1 a)2 -(cos-16)2 +(sec~1c)2 -(cosec-1d)2, , 5, , Sol. (a) Z sin-1(sin(2n-l)), , _2, , n=l, , =sin-1 (sin 1) + sin-1(sin3) + sin-1 (sin5), , + sin-1(sin7) + sin-1(sin9), = 1 + It-3 + 5-271 + 7-271 + 371 -9 = 1, • Ex. 9. Ifa and^(a > P) are roots of the equation, , (c)^, , (d)2, , (a)T, , (c)0, , (d)~, 2, , 7t2, Sol. (c) As 0<(sin-1a)2 <—,0<(cos 1 b)2 <7t2,, 4, 2\, , 2, , 0<(sec-1c)2 ^7t2 except— and 0 <(cosec-1d)2 <—, I, 4J, 44, , (cos-1 a + tan-1 a + tan-1 p) is equal to, , ... 571, , (b)-y, , (, , x2 - 41x + ^2 - 2V2 = 0, then the value of, , z \ ^7t, , (a) - it2, , (b)T, , 8, 3, Sol. (a) x2 ~^2x +-^3-2^2 =0 => x.22-fix+J2-1 = 0, , So, 0 <(sin-1 a)2 + (cos-1 b)2 + (sec-1 c)2+(cosec-1d)z ---2, *, 5%^, .•.(sin-1 a)2 + (cos-1 b)2 + (sec-1 c)2 + (cosec-,d)2 =---- (Given), 2, / . -i x2 n2, =>, (sin a) = —, , => x2-l-V2(x-l) = 0 =» (x-l)(x+l-^) = 0, , (cos-1 b)2 =7t2, , x = l,f2-l, , (sec-1c)2 =7t2, , www.jeebooks.in, a = 1 and p = Ji -1, , r., , -1, , -1, , -in, , 3jt, , Hence, cos a + tan a + tan p = 0 + — + — = —, 4 8, 8, , and, , (cosec-1d)2 =^-, , Hence, (sin-1 a)2 -(cos-1 b)2 + (sec-1 b)2 -(cosec-1d)2 = 0

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www.jeebooks.in, !, , Chap 04 Inverse Trigonometric Functions 347, , /, , (a) XG (tan2, tan3), , _______ 1_________, , • Ex. 12. Iff(x) = S tan, , lx2 + (2r-l)x+ (r2 -r+1)y, , r=l, , then lim f'(Q) is, , (b) xg (cot3, cot 2), , (c) x G(- °°, tan2)u(tan3, oo), (d) xg (- oo, cot3)u (cot 2, oo), , I, , I, , Sol. (b) Given,, (a) 1, , i, , (b)2, , (c)3, , i, , (, , kA, , (d)4, n, , (O.rt), , y=K, , /(x) = tan ’(x+n)-tan-1(x), 1, , 1, , l + (x + n)z, , 1 + x2, , =>, , y=3, , y=2, , no)= Al + n2, , "(cot3,0) (cot2, 0) 0(0, 0), , Now, lim/'(0) = -l=> lim/'(0) =1, , -----y=0, , it, tan 1 x + 2 - — > 0, => cot x l-i, tan x + n2 —, 2J, k, 2, (cot-1 x -3) (2 - cot-1 x) > 0, =>, , • Ex. 13. The range of the function, , -i, , =>, =>, , (b)[——, 12, , .. (. 3n, (c) 0,—, k 4, , 2, , n, , -j, , As tan x— = -cot x, k, 2, (cot"’ x-3)(cot-1 x-2)<0, , /(x) = sec-1 (x) + tan-1(x), is, , (a)(0,7t), , 71 i, , I, , 4, , So/, (a) /(x) = S(tan-1(x + r) - tan-1(x + r -1)), , =>, , 1, , (cot- x)(tan x)+ 2---- cot x-3 tan x-3 2---- >0, I 2J, \, 2J, , 2<cot-1x<3, , => cot3 <x<cot2 (As cot-1 x is a decreasing function), , (d) None of these, , Hence, xg (cot 3, cot 2), , Sol. (a) Df =(-ooi-i]u[l,°°), , • Ex. 15. Let /(x) = sin(sin"12x)+cosec(cosec"12x), , Also, f is an increasing function., 3n, For,, x G(/(x)g(o. ,—, 4 ., , and for, , xg[1,~)J(x)g, , 7t, , —, , 4, , J, , + tan(tan-1 2x), then which one of the following statement, , (0, (ii), , .‘.For range of f(x),(i)u(ii) => (0,7t), • Ex. 14. The solution set of inequality (cot 1 x) (tan 1 x) +, , 2-—j cot, 2J, , (a) /(x) is odd function, , (b) /(x) is injective, (c) Range of /(x) contains only two integers., , (d) The value of f'\, , is equal to 6., , (, , i, , x-3tan, , is/are incorrect?, , x -3 2-------- >0, is, , I, , 2j, , Sol. (d) Clearly f(x)=6x and domain =•, , JEE Type Solved Examples:, More than One Correct Option Type Questions, • Ex. 16. If /(x) =cos 1(cos(x +1)) and, g(x) = sin~1(sin(x + 2)), then, (a)/(1) + £(1) = (n-1), , (b)/(1)>^(1), , g(l)=sin-1(sin3)=(tt-3), , /(l) + £(l)=(n-l), and /(l)>g(l), /(2) = cos-1(cos3)=3, , www.jeebooks.in, (c)/(2)><g(2), , (d)/(2)<^(2), , Sol. (a, b, c) /(l) = cos (cos2)=2, , g(3) = sin-1 (sin 4) =(7t - 4)

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www.jeebooks.in, 348, , Textbook of Trigonometry, , oo, , 2, , • Ex. 17. E tan, , is, , 1, = - cot tan-1 - + cot (tan-13) + tan tan, , <('- + 2)2y, , r=1, , 3, , ,, ,, n, (a) 7t - (tan-12 + tan- 3) (b) —, , \, , ' ’ 4, , lit, , (d) % + (tan-12 + tan-13), , (C)T, , r, , Sol. (a, b)Tr = tan, , (___, 3”3 11, , ' (r + 3)-(r + l)', , ____ 2, = tan, k(r2 + 4r + 3) + l>, , J+(r + 3)(r + l),, , Tr = tan-1 (r + 3) - tan-1(r +1), , 1 + 3.37/, , 4, 3, , „ 1 4, = -3 + - + - =, 3 3, , 4, LtanA = FI tan A = —, 3, 41, „ 1 . 1 4 . 4,, _ 4, E tan A tan B =-3.- + -.- + -(-3) = -5 + - =, ’ 3 3 3, 3' ■', 9, 9, sin2 (A + B) + cos2 C =sin2(7t - C) + cos2 C = 1, , Sum = 71 -(tan-,(2) + an"'(3)), ., , ., , .., , • Ex. 19. Which of the following is/are correct?, , 71, , = 7t-(7t + tan (-1)) = —, 4, , (a), cos(cos(cos-11)) <sin(sin-1(sin(7t -1))) <sin(cos-1(cos(2rt -2))), , • Ex. 18. If sides AB, BC and CA of a triangle ABC are, represented by x + 2 = 0,3x + y = 0 and x + 3y + 2 = 0, respectively, then identify the correct statement., 4, (a) ZtanA = —, , (b) cos(cos(cos-11)) <sin(cos-1(cos(27t -2))) <sin(sin-1, , (sin(7l -1))) < tan(cot-1(cot 1)), 5000, , 2500, , (c) X cos-1(cos(2t7i -1)) = X cot-1 (cot(t7t + 2)), where tel, (d) cot-1cot cosec-Icosec sec-1 sec tan tan-1cos, , (b) ntanA = -y, , cos" 'sin- 'sin 4 = 4 - rt, 41, , (c) EtanAtanB = ——, , Sol. (a,b,c,d) For (a) and (b), cos(cos-11) = 1 => cos(cos(cos-11)) = cost, , (d) sin2(A + 8) + cos2 C = —, 4, , sin-1(sin(7t-l)) = 7t-(n -1) = 1, , sin(sin-1 (sin(7t -1))) = sin 1, , Sol. (b, c), 71, , ,„, , cos-1(cos(2ti, , 1, , ZA = — + tan 2, 3, , sin(cos-1(cos(27t -2))) =sin2, , ZB = — - tan-13, 2, ZC = 7t-(A + B), = tan-13 -tan, , tan(cot-1(cot 1)) = tanl, It is easy to compare cosl,sinl,sin2, tanl, cosl <sinl <sin2 < tanl => (a) is correct, For (c), v cos-’cosx is periodic with period 271, .‘., cos-'cos(2tn-l) = cos-1(cosl) = l (tel), , 1, 3, , =2 tan-13- —, 2, Y, , 5000, , \B, , X cos-1cos(2t7t -l) = 5000, , 'i, , n— tan', , X'-, , A, , -2))=cos-l(cos2) = 2, , Now, cot-1 cot (tn + 2) =2 [cof'cot x is periodic with period n], 2500, , n~ tan-13, •X, , :. X cot-1cot(t7t + 2)=5000 => (c) is correct, , (d) sin-1sin4 = 7t -4, , cos cos-1 (7t - 4) = 4 - 71, tan tan-1 (4 - 7t) = 7t - 4, , tanA + tanB + tan C = tan I — + tan, \2, , n, , ---- tan 3, 2, , 1, + tan tan 3-tan', 3, , sec-1 sec (n - 4) = 4 - 7t, cosec“'cosec (4-7t) = 4-71, , cot-1cot(4 - 7t) = 4 - 7t, , www.jeebooks.in, => (d) is correct

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 349, , • Ex. 20. Let x, and x2 (xt > x2) be roots of the equation, -1, , sin (cos(tan, I, , / x, , -1, , (cosec(cot, , -1, , sin (cos(tan-1 -Jl + x2)) = —, 6, , ----- = 7C, *2, , 1', sin — + cos, (b) sin"'I, , 1, , sin, , 1*2,, , =0, , fx2 + 2=2, x2=2, , =>, =0, ,*2>, , + cos, *i, , n, 6, , x2 + 2, , (c) sin-1— + sin, , (d) cos, , 7T, , x))))=— then, 6, , 1, , ■ -1 1, , (a) sin — + cos, , *i, , Sol. (a,c,d) Given, sin-1 (cos(tan~’(cosec(cot-1 x)))) = —, 6, , = 7t, <*2, , =>, , x = ±f2, , So,, , X] =4iand x2 = -4i, , Now, verify alternatives., , JEE Type Solved Examples:, Passage Based Questions, Passage I, , 21. (h) Ash(x) = x,VxeR, k(x) = 1 + l(cos’* x + cot-1 x), , (Ex. Nos. 21 to 22), Suppose /, g and h be three real valued function defined, on R., , Letf(x) = 2x + |x|, g(x) = |(2x - |x|) andh(x) = /(g(x)), , 71, , Domain of k(x) = [-1,1] and fc(x) is decreasing function on [-1,1]., As fc(x) is continuous function on [-1,1]., Now,, , A:mi„ (x = 1) = 1 + — (cos-11 + cot-11), 7t, , • Ex. 21. The range of the function k(x) =1 + —, it, (cos-1(/j(x)) +cot \h(x))) is equal to, , . J1 7, , (a), , 4 4, , (b) 77, 4 4, , , J1 5, , (d)f-,—, , (c), , 4 4, , 4 4, , • Ex. 22. The domain of definition of the function, /(x) = sin-1(/(x) -g(x)) is equal to, , ft] , 1 5, 7t, . . 1if, = 1 + — 0 + — =1 + - = 4, 4 4, 71, (x=-1) = 1 + — (cos' ’(-1) + cot"’(-1)), n, 3tc, .= 1- +- —, 11 f 71 +—, 4, 7t, , =1 +1=“, 4 4, x F5 11, => Range of k(x)=, 22. (d) We have, /(x) - g(x) =(2x+1 x|) -1 (2x -1 x|), , (a)g.», , (b) (-«,!], , 4x 4, =— + - x, 3 3, , (c)[-H], , 3, (d) - °°>8, , lx; x>0, =• 3, 0 ; x<0, , Sol. (Ex. Nos. 21 to 22), x, ', 3x, > 0 , , . - ,x£0, We have /(x) =, „ and p(x) 3, x,x<0, ’, x,x<0, , Clearly, f and g are inverse of each other., , For domain of function, 0<—<1 => 0^x<l, 3, 8, i, , Domain of/(x)=l, , 3, , 31 — |=x,x>0, , www.jeebooks.in, Now, h(x) = f(g(x)) =, , x , x<0, , (Note Range of function /(x) = 0, y )

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www.jeebooks.in, 350, , Textbook of Trigonometry, , Passage II, , = tan', , (Ex. Nos. 23 to 24), , In AABC, if ZB = sec, , ZC = cosec, , l-’.i, , +cosec, , 5,, , 25, , <9, , + cot, , 7, , (b, , 4 2, -1^25^, -if9 A, ZC = cosec 11—J + cot I —I, , and c = 3, = tan, , (All symbols used have their usual meaning in a triangle.), On the basis of above information, answer the following, questions., , (b) GP, (d) neither AP, GP nor HP, , • Ex. 24. The distance between orthocentre and centroid of, 4, , triangle with sides a2, b3 andc is equal to, , <4, , (b)|, , 3, , 2, , = tan, , _, , _/13^, , I9J, , 124J, f 7 13 'I, —+—, 24, , = tan, , 9, , !_2_ 12, , = tan3, , 24 9 ,, , tanB =2 and tanC =3 => tan A = 1, v, EtanA = IItanA, 1 . „ 2, . . 3, .'., sin A =-j=,sinB = —j= andsinC = -?=, V2, V5, V10, a, b, c, —----- zz — ;z--------sin A sinB sinC, =s>, , 2, 3, Hence, a = V5 and b =2y/2, c =3, , 23. (a) tan A = 1, tanB =2, tanC=3 are in AP, 4, , Sol. (Ex. Nos. 23 to 24), , ZB = sec, , (7>, , <, , • Ex. 23. tan A tanB, tanC are in, , (a) AP, (c) HP, , 3 1, _4__ 2_ = tan 2, , -+-, , 5, + cosec, 4,, '3, , ,4, , 5, , 1, , + tan', , 2, , 24. (b) The triangle sides a2, b 3 and c will have side-length, 5, 4 and 3 respectively., Distance between orthocentre and centroid, 2 . ., . hypotenuse 5, = -(circumradius) =, ------ = -, , JEE Type Solved Examples:, Integer Answer Type Questions, 2, , • Ex. 25. Let /(x) = x — lax + a - 2 and, , g(x)= 2 +sin, , 2x, , . If the set of real values of ‘a’for, , 1 + x2, , which f(g(x)) <0,V xeR is (kv k2), then find the value of, (IO*, -3k2)., [Note : [fc] denotes greatest integer less than or equal to fc.], SoL (8) We have g(x) =2 + sin', , 2x, 1 + x2, , = 2 + sin', , 2x, 1 + x2, , As,, , 2x, sin1 -----1 + x2, , —re n, 2*2., , Range of g(x) = {0,1,2,3} for /(g(x)) < 0V xe R, =>, /(0)<0 and(3)<0, Now,, /(0) < 0 =>a — 2 < 0 => a <2, and, /(3)<0=>9-6a + a-2<0, 7, a>5, f(x)=x2-2ax-a-2, , 0, , 4-, , 1, , 2, , 4-, , 3, , -+, , --x-axis, , ae^|,2j, Hence,, , www.jeebooks.in, sin, , 2x, , (lOfcj-3k2) = 14-6 =8, , = -2,-1,0,1

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www.jeebooks.in, 352, , Textbook of Trigonometry, , ■■, , 71, , 2, , • Ex. 32. If tan, , f(x), , n, , sin, , 7t, 2, , 4, , n, "2, , x, , Vi+x 2 j, , -X, , 7n2jt, 4, , 0, , y - cos, , A, , — = — and, 3, , x + tan, , 7t, 5sin, X, =—, then-----is, 6, sin" y, 71, X=—, , Sol. (6) tan-1 x + cos-1 y = y and sin-1 y - cot, , 6, , g(x), , 71, , tan-1 x + cos-1 y + sin-1 y - cot x+2, , :. Required area = area of shaded square, _9n2 _ant2, ~ 8 ~ b, Hence,, , or, , tan x = cot- x, , Also, tan 1 x+cos-1y-sin-1y + cot, , a =9 and & =8, a-b = l, , it, , =» 7t-2sin, , • Ex. 31. Consider the curve y = tan, , I, , curve for i = 1,2,3,. ..n(neN) such thatyr = E tan, m=1, , 1, , ], , 2m2), , and B(x,y) be the limiting position of variable point Pn as, n—>°o, then the value of reciprocal of the slope of AB will be, n, i, Sol. (2) y - Limyn = Lim E tan’, n—>“, n—>“m = l, 2m2, Lim E tan, , n—»“m = l, , . -j, , • Ex. 33. If A = -cot, , L, , 1, 2" + -cot, , + —cot, 2, , 2>, , an c, — + —cot (3), b, d, wherea,b,c,dE N are in their lowest form, find (b- a- c-d), cof, , 1, , + - cot, 3, 2, , =2(cot-12 + cot-13)+ cot-13, , n, , ■OH, , n—, , = Lim {(tan-13 - tan-11) + (tan-15 - tan-13), It, , = — + cot, 2, , + (tan-17-tan 15)....4-tan 1, , 1, , 71, , = Lim {tan-1(2n + 1)- tan-11} —> —, n —°°, , 1, - + cot, 2, , 1, , 3, , n, , 1, + —cot, 6, 3, 2;, 1, , 3 - f — + - tan-12, [_4, , 6, , 71, _ i r 3tc, _i., = — + cot 3—----- tan 3, 4, 4, 6, , (2n + 1) - tan-1(2n -1)}, i, , - and, , 3, , B=1cot-1(1) +2cot-1(2) +3cot-1(3) then |B-/4| is equal to, , Sol. (8) B - A =(2cot-1(2) + 3cot-1(3)) -, , _______ 2, 1 + (2m+ l)(2m-l), , It, , 4, , = — + cot 3 + -tan-13, 8, 6, 7t, 1 ( it, n, = — + cot 3—---- cot 3, 6<2, 8, , i.e. coordinates of B approach, towards those, , of ‘A’., , .-. Chord AB approches to be the tangent to y = f(x) at A, , 7t, , 71, , 571, , 5, , = — + — + cot 3--cot, 6, 8, 12, , i rd, .'.(slope of AB) = —tan x, dx, Jat x=l, =(l + x2)I=I = 2, , 6, , 571, , sin y = —, 12, , = Lim E {tan-1(2m+l)-tan-1(2m-l)}, , ■;B^>, , 71, , x=—, , 5sin~ -^=6, sin-1 y, , x and a point, , am on it. If the variable point Pj {xhyt) moves on the, , n, , x=l, , Hence,, , 3, , _1(, , = — + - cot .3, 24 6, a =5;b = 24;c = 5;d =6, b-a-c-d=8, , www.jeebooks.in

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , 353, , JEE Type Solved Examples:, Statement I and II Type Questions, , • This section contains 2 questions. Each question contains, Statement I (Assertion) and Statement II (Reason). Each, question has 4 choices (a), (b), (c) and (d) out of which only one, is correct. The choice are, (a) Both Statement I and Statement II are correct and, Statement II is the correct explanation of Statement I, , x, y, z is equal to 1/3., For any x, y, z we have, , Then, , A + B + C=4, x + y + z-xyz, Now, tan(A + B + C) =, l-(xy + yz + zx), , a exists but not cos ”10 (a > 0)., , 1 - x + y + z~xyz, l-(xy + yz + zx), , Statement II Domain of cos-1 x /s[-1,1]., , -1, , -3, , l-(xy + yz + zx) = x + y + z-xyz, =>, (x-l)(y-l)(z-l) = 0, =>, => One of x,y,z is equal to 1, z=l, x + y = 0, If, (x)odd + (-x)°dd + 1*^ = 1, , 3, , 2 ,, , Thus, AM of odd powers of x,y,z is equal to, , Sol. (a) Given, 6x2 + llx + 3 = 0 => 6x2+ 9x + 2x + 3 = 0, , 3x(2x + 3) + l(2x + 3) = 0, , -3 -1, 2 ’ 3, , (2x + 3)(3x+l) = 0 => x =—,—, , „ -3, -1, P=, —, a = —, , ■: — > —, , 3, , _/-l, 3, , _•, , 7t, , • Ex. 34. Statement I Ifcl, 0 are roots of6x2 +11x4-3=0,, , cos, , x, , Sol. (b) We have, tan”1 x+tan”1 y + tan z = —, 4, Let x = tan A,y = tanB and z = tanC., , (d) Statement II is correct but Statement I is incorrect., , 2, , Jt, , x + tan y = — tan 2, 4, , xyz -xy-yz-zx + x+y + z =1 + (x-1)(y-1)(z-1), , (c) Statement I is correct but Statement II is incorrect., , =>, , If tan, , and x+y + z = 1, then arithmetic mean of odd powers of, Statement 11, , (b) Both Statement I and Statement II are correct but, Statement II is not the correct explanation of, Statement I., , then cos, , • Ex. 35. Statement I, , j., , {•/ Domain of cos”1 x is [-1,1]}, , exists, , JEE Type Solved Examples:, Matching Type Questions, • Ex. 36. Match the principal values ofcos “1(8x4-8x2 +1), , (B) 0 <cos-1(8x4 ~8x2 + 1) <71, , given in column I with the corresponding intervals ofx given, , =>, , 271 <4cos-1x<37t, 71, -1^.371, 1 . .., — <cos x<— => —j=<X<0, 2, 4, 42, , in column II, for which it holds., , Column II, , Column I, , A, , 4 cos” x, , P-, , 0<x<-L, V2, , B, , C, , 4 cos” x-2n, , qr., , 2n - 4 cos” x, , (C) 0 <2?r - 4cos”'X <7t, =>, , - 271 <-4cos_1xS-7t, , V2, , =>, , 27t>4cos”1x£7t, , -i<x<-4=, , =>, , — >cos x>— =>, 2, 4, , --l=<x<0, , (D) 0 47t - 4 cos" x < 7t, -4ti <-4cos”’ x^-37t, =>, , -J=<x<l, 42, , D, , s., , 4ti-4 cos x, , 0<4cos”lx-27t <7t, , 42, , 7t, , -1, , 71, , 0<x<4=, V2, , 4rt >4cos”’x>37t, ., -1 ^3n, 71 > COS X >-----, , www.jeebooks.in, =>, , (A) 0 < cos-1(8x4 - 8x2 + 1) < 7t, , =>, , 0<4cos-1x<7r, i, , 7t, , 1, , 0<cos x < — =>, <x< 1, 4, 42, , 4, , -1<x£-4=, 42

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www.jeebooks.in, 354, , Textbook of Trigonometry, , Subjective Type Examples, , • Ex. 37. If A = 2 tan 1(2>/2 -1) and, B = 3sin, , =>, , n + sin, , 2(tan"’ x)2 -7C tan, , 3 I, - , then show A> B., 5J, , ‘, , n, , 7t 371, , tan x =---- ,—, 4 4, , Sol. We have,, , -1, , =>, , A=2tan-1(2V2-l) = 2tan"’(1.828), , n, , tan x =---4, , A >2tan“’(/3), , x = -l, , 271, , A> —, 3, , =>, , ZTt2, 8, , x------- = 0, , 7t n, 2’2, , 371, ,, ., -1, JU, neglecting tan x = — as tan xe, 4, , —(i), , Also we have,, , sin, , 3 sin, , n, 3;, , fi, , 1, , => sin, , 2, , 3, , 71, , • Ex. 39. Solve forx: //[sin Vos 1sin Van 1 x] =1, where, , <—, 6, , [.] denotes the greatest integer function., , 71, , <—, 2, , 1, , 3sin, , Also,, , 1, , <sin, , Sol. We have, [sin“1cos"1sin"1tan"1 x] = 1, , 3---4| 3 <3,, , =sin, , 3, , n, 1 <sin"’. cos"’ .sin"’. tan x< —, 2, sinl < cos-1 .sin-1. tan' x<,l, , J, ^)=sin"’(0.852), , =sin, , =>, , cos sin 1 > sin"’.tan"1 x>cosl, , sin cos sin 1 £ tan"1 x > sin cos 1, , 1, , 3sin, , tan sin cos sin 1 > x > tan sin cos 1, Hence, xeftansincosl, tansincossinl], , 3 1, , 3, 5, , I2, , tit, , of the equation x4 -6x2 +1=0., , 3, , + sin, , 3, , 271, , 5, , 3, , •••(ii), , Sol. We have,, , tan~’y = 4tan~’x, 2x, 1-x2, , => tan y =2 tan', , 1-x2, = tan —, 4x2, 1(1-x2)2, , 8, 57t2, , Sol. We have (tan-1 x)2 + (cot"1 x)2 =---8, =❖ (tan-1x+ cot"’x)2-2 tan"1 x. cot, , 5tc2, x =-----, , I ---71 tan, » -1x, -2 tan' x., k2, v tan x+ cot, n., , 71, , x = — =^cot, 2, -i, , = tan', , 8, , 2, , .2, , 57t2, , ----- 2 — tan x + 2(tan x) =----4, 2, 8, , 5tc2, , 8, 71, , x =---- tan x, 2, , (as | x| <1), , 4x, , 2, , C, , _i, , ., , 8, , 7t, , • Ex. 38. Solve for x .-(tan-1 x)2 +(cot-1 x)2 =------ ., , „ 71, , TC ., , algebraic function ofx and hence prove that tan— is a root, , <—, , B=3sin, , Tt2, , * | x | < tan — , findy as an, 8J, , 1, , From (i) and (ii), we have, A>B., , =>, , 71 I, , • Ex. 40. //tan“1y = 4tan, , - I=sin” (0.6)<sin', , sin', , Hence,, , =>, , 1, it, <—, 3, 3, , sin, , =>, , <sin, , 3, , 3sin, Also,, , 73, 2, , =>, , if, , 4x(l-x2), , as, , | 2x |, "-J<1, , I Ml ., , x4 -6xz +1, , 4x(l-x2), , y, , x4 -6x2 + 1, 7t, , x = tan—, 8, 7C, , tan" y = 4 tan' x = —, 2, , www.jeebooks.in, y = 00, , => x4 -6x2 + 1 = 0

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www.jeebooks.in, 358, , Textbook of Trigonometry, , • Ex. 52. Solve:, , • Ex. 53. Obtain the integral values of p for which the, , 2x, , 3sin, , 1-x, , 2, , — + 2 tan, 1 + x2, , - 4cos, , 1 + x2, , 2x, , 71, , 1-x2, , 3, , Sol. Let tan-1x = 0for x>0, 71, , 71, , Case I When 0<x<l, then 0 <0< — and so O<20< —, 4, 2, 2x, sin', = sin 1(sin20) = 20 = 2tan-1 x, 1 + x2, , 1-x2, cos *------ 7 = cos (cos20) = 20 = 2tan x, 1 + xz, 2x, tan, = tan-1(tan20) = 20 = 2 tan-1 x, 1-x2, f, l-x2>, 2x, 2x, -4 cos, + 2 tan', + x2y, 1 + x2, 1-x2,, \, , /, Thus, 3 sin, , following system of equations possesses real solutions :, 77 4, , rm2, , x + (sin-1y)2 =^-— and(cos~' x)(sin-1 y)2 = —, 16, 4, Also, find these solution., , cos, , Sol. Let cos-1 x = a =>a G[0,7t], , and, , sin-1 y = b => b e, , We have, , a+b, , -.(i), , nh2-^, ab = —, 16, _2, , and, 7t, , Since, , i.e., , 0<p<—+1, 71, , 3, , Since, p 6 Z, so p = 0,1 or 2, But, if p = 0, then a = b = 0., => Equation (ii) will not be satisfied., Now, substituting the value of b2 from Eq. (i) in the Eq. (ii), we, get, , 7t, 1, X = — => X = —7=, 6, y/3, , 0<-4=<l, x=-Lis a solution, v3, V3, 2x, Case II When x = 1, tan', is not defined., 1-x2, , as, , a, , x = 1, cannot be a solution., 71, , 71, , p7t2, , I 4, , J, , Since,, , Case III If x > 1, then — < 0 < — and so — < 20 < 7t, 4, 2, 2, , and tan, , 2x, =sin"1(sin2G) = 7t-20 = 71 -2 tan x, + x 2y, 2x, = cos"’(cos20) =20 = 2 tan-1 x, 1 + x2,, , 2x, = tan ’(tan20) =20-71 =2 tan X —7t, ,1-x2, , = ^- =>16a2-4p7t2a + 7t4 =0, , i.e. 16p27t4 — 64ti4 >0 => p2>4=>p>2, Thus, we conclude that the only value of p that satisfies all, conditions is p = 2 . Substituting p = 2 in Eq. (iii), we get, 16a2-87t2a + 7t4 =0, , (4a-7t2)2 = 0, 7t2, , =>, , «, , 7t -10 tan x = —, 3, , _.4, , _2, , From Eq. (ii), we get---- b2 = — =>b = ± — =sin-1y, 4, 16, 2, =>, y = ±l., , 7t, X =—, 15, , • Ex. 54. Solve the equation 2(sin 1 x)2 -(sin-1 x) -6=0, , 7t, , Sol. Let, sin-1 x = y, we get, , or, , tan, , i.e., , x = tan— <tan—<1, 15, 4, , 71, , x = cos—, 4, , 71, , 3(71-2tan x)-4(2 tan x) + 2(2tan x-7t) = —, 3, 1, , _j, , a = — = cos x, 4, 7t2, , 1, , 1, , ...(iii), , aeR => D>0, , Thus, the given equation becomes, 1, , 4, , 4, , 7t2, pit2, So, Eq. (i)=>0<-^—<7t + —, 4, 4, , 7t, , cos, , 7t2, , b2 e 0, —— =>a + b2e 0,71 +----, , 3, , 2 tan X = —, , sin, , 2’2., , 4, , => 3 (2 tan" x) - 4(2 tan-1 x) + 2(2 tan-1 x) = —., 3, , tan, , _7t 7t, , x = tan — is not a solution., 15, , 2y2—y —6 = 0, , 2y2~4y + 3y-6 = 0, y =2 and y = -1.5, sin-1 x=2 and sin-1 x = -1.5, , www.jeebooks.in, 1, Thus, x = —j= is the only solution for the given equation for, 3, x£0., , 7t, , Since 2 >— and |-1.51 <, , 71, , the only solution is x=sin(-15).

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 359, , • Ex. 55. Solve the equation sin-16x + sin-1 6^3% - -----71, 2, Sol. Let us transfer sin'16 Vix into the right hand side of the, equation and calculate the sine of the both sides of the, resulting equation, , • Ex. 56. Solve the equation : 2 tan \2x -1) =cos, , 2tan-1(2x-l) = cos-1x, , Sol. Here,, , cos(2 tan-I(2x -1)) = x, , or, , ,, n l-tan20, We know cos20 =-------- —, l + tan20, , l-(2x-l)2, l + (2x-l)2, , sin(sin-16x) =sin| -sin-16-Ax - — |, , 2J, , \, , x., , 6x = ~sin(sin-16^3x + sin-11), , {using sin(-0) = -sin(0)}, 6x = -sin(sin-1 -^1 -108x2), , -2x2 + 2x, = x => 2x3-x = 0, l-2x + 2x2, , =>, , _ V2 ^2, x = 0,—,----- ., 2, 2, Now, to verify which of the following satisfy given equation,, Case I For x = 0;, 2 tan“’(-l) = cos-1(0), , =>, , {using sin-1x+sin-1y =sin-1{x-Jl -y 2 + y^jl - x ’ll, , 6x = -71-108x2, , ...(i), , Squaring both sides, we get, 36x2 =1 -108x2 => 144x2=1, ,, 1,1, whose roots are x = — and x =---- ., 12, 12, Let us verify:, , -Ji, , j/T, , + sin, , 2>, , n, 6, , —, 3, , n, , z——, , substituting x = — in Eq. (i), we get, 12, , RHS, , Case II For x = —, 2, .The right hand and the left hand side of the equation are equal, , to — and 2 tan-1(V2 -1)., 4, tan’’(V2 -1) = —., 8, , But, , 2, , Thus, x=-^ is the root of given equation. But, when, , LHS, , n, , 2 “2, , x = 0 is not a solution of given equation., , Substituting x = — in the given equation, we get, sin, , n, , or, , 6x = l, 2, -71-108X2 =-1/2, , i.e. LHS* RHS of Eq. (i)., Hence, x = —— is a root of the given equation as it satisfy both, 12, given and Eq. (i)., , ■v2, , x = — is a root of the given equation., 2, V2, Case III For x =----2, The left hand side of the equation is negative and the right, hand side is positive., , •J2, , Consequently, x = —— is not a root of the given equation., , Thus from above;, ■^2, , x= — is the only solution., , www.jeebooks.in

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www.jeebooks.in, 0 Inverse Trigonometric Functions Exercise 1:, Single Option Correct Type Questions, , 1, , 1. cot-1, 1 + x2, , /, , (a) cos"1 (x2), , (b)^-|cos-1(x2), 2 2, , (c)", J2 cos~’ (*2), 3, , (d) None of these, , (1, -i1)1., 2. The value of co:•a -cos - I is equal to, \2, , (b) x = 2 + ^9-2n, -271, 2 + ^9 - 2n), , 4. The value of sin -1 • I sin —, , I, , 3 7(x2 + k2 - kx), , K, , x, , 6, , 7(x2 + k2 - kx), , COS —, , x, , «, where, , (a) (cot 3, cot 2), (b) (- 00, cot 3) u (cot 2, 00), , 9. Sum to infinite terms of the series tan, , /, , x2 + 2xk-2k2>, , ^1 + 2, , 1, 10. If x + — = 2, the principal value of sin'-1 x is, x, , / '71, , (b)i, , (c)7t, , (d)^, , (b)2x, (d)x, , <b>?, , (d) None of the above, -/1-x, , (a) 0.71, n ’7n, <(c>\[“7, \, , 4 4, , + ..., , 2n, 12. If sin ’x + sin-1 y = —, then cos 1 x + cos 1 y, 3, , x2 + 2xk - 2k2, 2x2 -2xk + 2k2, , 11 + x, , 2n-l, , -if 2, , (b)n/2, (d) None of these, , (a) x/2, (c)3x, , x2 - 2xk + k2 /, , 5. If a < tan, , 2"-1, , + ... + tan 1, , + tan, , 11. Ifxel--,- then the value of, 2 2, ' 3sin2x, tanxA, tan-1, + tan-1, is, + 3 cos 2x, 4 J, , ! 2x2 + xk kx2 -2xk + k2 ), , (c) tan-1, , 8. Solution set of the inequality, (cot-1 x)2 -(5cot-1 x) + 6>0is, , <a)7, , k, 1, — < x < 2k, k > 0 is, 2, ), , (b) tan, , (d) None of these, , (a) tc/4, (c)7t, , (d) x > 2 + ^9-271, , (a) tan, , (c) x = 0, , +,an"fe), , (a) x = 2 - -^9 -2n, , - cos-1, , (b)x = l/V5, , (d) None of the above, , 3. The inequality sin "1 (sin 5) > x 2 - 4x holds if, , (c) x 6 (2 -, , (a)x = 3, , (c) (cot 2, 00), , (b)-3/4, (d) 1/4, , (a) 3/4, (c) 1/16, , 71, , 7. Solution of equation cot -1 x + sin -1 — = — is, , is equal to, , (d)n, , < b where 0 < x < 1, then (a, b) =, , (b) \ °.yM J, , <d)(H, , 6. Sum of infinite terms of the series, 3, cot-1 l2 +2| + Cot-,[22 +- + cot-1|32 +- + ... is, 4, 4, I, 4, , 13. cot| — -2 cot 1 3 is, , 14, , (b)7, (d) None of these, , (a)l, (c)-l, 14. sin tan 1, , 1-x2, 2x, , (a) 1, (c)-l, , + cos, , -U-y2, 1, , 1, , „2, , is, , (b)0, (d) None of these, , www.jeebooks.in, (a) n/4, (c)tan-13, , (b) tan-12, (d) tan-1 4

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 361, , 15. If cos, , = 2 tan 1 x, then x is, , -cos, , 1 + b 2,, , ,1 + a2,, , O’—, 1 + abA, , 1 + ab, .. a + b, (c)------l~ab, , (d) None of these, , f, , 16. If, , cos, , (c)xe, , It, , 24. If sin-1 : [-1,1] —> —, — and cos-1 : [- L1] —>[0, it] be, , 1, , two bijective function, respectively inverses of bijective, ., 7t 3rt, functions sin : —, — -i [-1,1] and cos: [0,7t] —» [-1,1] ,, .2 2 ., , < —, then, 3, , [1 + x2,, _JL _L, , (a)xe, , (b)xe -, , L0>~r, 41], , then sin-1 x + cos-1 x is, , \n, , (d) None of these, , (b)n, , t \3n, , 17. The value of, /, 4-cos-1, , (a)0, , (b)74, , (c)76, , (d)72, , IT + sec, 14,, , 1, , X 7t, 18. If tan"1 — < —, x g N, then the maximum value of x is, 3, , (a) 2 ., (c)7, , (b)5, (d) None of these, , 19. If tan-1 ^(1+_y2) VqZZ— = a, then x2 is, 2), , 7(1 + x2) + a/(1- x 2), (a) cos2a, (c) tan 2a, , (d) not a constant, , CT, , f, , cos-1 cot sin-1 ., , 7C, , where [.] denotes the greatest integer function, is equal, to, (a) (0, cotl), (b)(0, tanl), (c)(tanl, oo), (d)(cotl, tanl), , L2 2 _, , .2 A, , 1-X, , _j, , 23. Complete solution set of [cot -1 x] + 2[tan 1 x] = 0,, , (b) sin2a, (d) cot 2a, , 25. If a sin-1 x-bcos 1 x = c, thenasin-1 x + bcos ’xis, equal to, , nab + c(b - a), a+b, nab + c(a — b), (d) (d)---------- -----a+b, , (a)0, , (b), , n, (c) 2, , 26. The number of integer x satisfying sin -11 x - 21 + cos 1, (l-|3-x|) = iis, , (b)2, (d)4, , (a) 1, (c)3, , 27. The value of a such that sin, , functions, then the maximum range of value of x is, ' n .IL n, fl", (a) —, -1 U 1,—, , 2, , J, , 2, , (b), (c) (-oo,, , tan-1 2x, , tan-1 3x, , tan-13x, , tan-1 x, , tan-1 2x = 0, then the, , tan-1 2x, , tan-13x, , tan-1 x, , number of values of x satisfying the equation is, (a) 1, (b) 2, (c) 3, (d) 4, , one real root a, then the value of 2 tan 1 (cosec a) +, , 22. The value of lim cos(tan- ’(sin(tan-1 x))) is equal to, , w-4-41, , 28. Let, , tan-1 x, , 29. If the equation x3 + bx2 + cx + l = 0,(b<c), has only, , °°), , (d) [-i, o) u(o, 1], , (a)-l, , Vio, , (b)l, , (b) two, (d) four, , 21. If cosec- ’(cosec x) and cosec(cosec- ’x) are equal, , 4s, , the angles of a triangle is, , 20. The number of positive integral solutions of, tan-1 x + cot-1 y = tan-13 is, (a) one, (c) three, , sin-1 ~^=, sin-1 a are, , -I, , (b) 41, , tan-’(2 sin a sec2 a)is, (a)-n, , (b)--, , <4, , (d)n, , www.jeebooks.in

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www.jeebooks.in, 362, , Textbook of Trigonometry, , 30. Let u = cot"1 Vcos 29 - tan 1 Vcos 29, then the value of, , sinu is, (a) cos29, (c) tan2 9, , (b) sin 29, (d) cot2 9, , 31. Let/(x) = 2tan”1 x =, , sin, , x>0, , kl + x2J’, , (a) log, , J + x2,, , ,, , continuous everywhere but not differentiable at x is/are, (a) 0,1, (b) -1,1, (c)-l, 0, (d)0,2, , 2 J, , (b) log, 7, , 34. Let /(x) = sin-1 2x + cos-1 2x + sec-1 2x. Then the sum, , 2 J, x -1, 1 - sin, 2, , I 2 ), , (c)e, , ,1 - sin, • -1 ( -----x ~ 11, , 2 >, , (d)e, , 2 >, , 39. cos ”1 (cos(2cot”1 (^2 -1))) is equal to, (a) y/2 -1, , (b) 4, , 3n, , (d) None of these, , cT, , 40. The maximum value of /(x) = tan 1, , ' (V12 - 2)x2, , (b) 36’, , (a) 18’, (c) 22.5’, , (d) 15’, , 1 -fi+s — = 4°, then, X, , (a) x = tan2°, (c) x = tan(l I 4)°, , (b)x = tan4°, (d) x = tan8°, X, , of the maximum and minimum values of /(x) is, (a) it, , (b) 2it, , (c)3K, , (d)|, , 35. If tan 1, , b, , — + tan, c+a, , _t, , c, , ------ = —, where a,b,c are the, a+b 4, , sides of AABC, then AABC is, (a) Acute-angled triangle, (b) Obtuse-angled triangle, (c) Right-angled triangle, (d) Equilateral triangle, , It, , 1, , 42. If tan-1(sin2 9-2sin9 + 3)+ cot-1(5 sec2 y +1) = -, then, 2, the value of cos2 9 - sin 9 is equal to, (a) 0, (b) -l, (c) 1, - „ —None-of the above, , 43. The number of solutions of the equation, | tan-11 x11 = -J(x2 +1)2 —4x2 is, (a) 1, (c) 3, , 36. Solutions of sin” ’(sin x) = sin x are, if x e (0, 2tc), , (b) 2, (d) 4, , 44. For any real number x > 1, the expression, sec2(tanx) - tan2 (sec-1 x) is equal to, , (a) 4 real roots, (b) 2 positive real roots, (c) 2 negative real roots, (d) 5 real roots, , (a)l, (c) 2x2, , (b) 2, (d) 2^2, , 45. Let f :R —> 0, ~ J be defined by, , 2 sin”1, ’, , 37. The equation e, , is, , 4 + 2x2 + 3y, , (b) [81, «), (d)(-00,81], , ■ I ------x~ 1 I, sin, , ■ f*"1), , 41. If tan, , (c)(-~, 81), , /, , --, , sin ------, , 33. Let /(x) = tan”1 (x2 - 18x + a)> 0 V x g R. Then the, value of a lies in, (a) (81, ~), , , x > 0, then f 1 (x) is equal to, , x<0, , V x e R. The function f(x) is, , J+x2J, , X, , <e* + 1J, , 2, , The function f(x) is continuous everywhere but not, differentiable at x equals to, (a)l, (b)-l, (C)O, (d)i, V2, 2x, , e, , . -1■ ( Xx — 11, 1 - sin, —■—, , 'l-x2>, -cos, , 32. Let/(x) = sin 1, , 38. Let f(x) = 1 + 2 sin, , (assuming f is bijective), x-n ), ( . -ifx-n, , -/l~x2: A, , cos, , /, , -, , n, , (a) Unique solution, (b) Infinite many solution, (c) x = l, (d) y = e, , y, , logy’, , has, /(x) = tan-1(3x2 + 6x + a). Iff(x) is an onto function,, , then the value of a is, (a)l, (c)3, , (b)2, (d)4, , www.jeebooks.in

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 363, , 46. The value of expression, 72, , tan, , 2, , V5, , 53. The value of, , cd, , + sin, , (a) cot, , (b) cot, 1 4- V2, , (c) - n + cot, , 1-V2y, , 47. The value of sec 2 cot, l \25, , 10 J, , (cot A) + tan, , -13, , (cot A) for, , (b)2tan ’(2), , (a) 4tan-l(l), , (d) None of these, , (c)0, /, , 4n, , 54. The sum E tan, . ., , 2, , -i 1, , is equal to, , -2n2 +2, , n=1, , -i 3), 2 + cos - I is equal to, , (b) 4tan”'l, , 24, , (a) tan - + tan, 2, 3, , (c)7, , (d) sec"'(—Ji), , 25, , (d)-y, , (C)y, , ,, , 0< A < (tc/4) is, , \/2 + l, , (b)-7, , i \25, , A, , 1, , - tan 2A + tan, 2, , tan, , is, , <72-1, ri -72^, (d) n - cot, j + Ji), , /LX, , (.)-, , VI0, , -cos, , 55. Number of solution(s) of the equations, , 48, Which one of the following statement is meaningless?, / x -i ft f2e + 4^, (a) cos I In I—-—II, (b) cosec, , (l-x)-2cos-1 x = —is, 2, (a) 3, (b) 2, (c)l, (d)0, , cos, , (d) sec-I(7t), , 49. The value of sec sin, , ., , 507C], , - sin----- + cos, 9 J, , co:, , 317C, , 9, , equal to, I \, , 10n, , (a) sec---9, (c)l, , (b)sec —, 9, (d)-l, , = k. The sum of all possible values of k is, , •, , (c)0, , 21, (b - —, 40., , 1, , 51. The value of E tan, r=2, , (b)i, , (C)^, , (d)^, , x, , /2 -5r + 7,, , 57. The range of values of p for which the equation, , sin cos-1(cos(tan-1 x)) = p has a solution is, , ()< l, , _L, , (cfe‘), , (d)l, oo, , (a)21, ,, , 50. The number k is such that tanfarc tan(2) + arc tan(20C)}, , / X 19, (a), 40, , is, , 56. There exists a positive real number x satisfying, ( ji A, X, cos (tan-1 x) = x. The number value of cos, is, , (b) [0.1), , (d)(-l.l), , 58. Number of solutions of the equation, , > is, , logiq (-Jscos-1 x -1) + - log10 (2cos-1 x + 3), 2, , i \ K, , ()7, ., , 3n, , CT, , <b)I, , + log10 V5 = 1 is, , (d)^, , (a) 0, (b) l, (c) more than one but finite, (d) infinite, , 4, , 52. If x = tan, , fl, 12, , 1 - cos, , y = cos - cos, , Bthen, , (a) x = izy, (b) y = nx, (c) tanx = -(4/3)y, (d) tan x = (4 / 3)y, , + sin, , 1, 2*, , 59. Which of the following is the solution set of the, equations sin-1 x = cos-1 x+sin-1(3x-2)?, , wfp1, , www.jeebooks.in

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www.jeebooks.in, 364, , Textbook of Trigonometry, , /, , 60. The set of values of x, satisfying the equation, tan2 (sin-1 x)> 1 is, , (a) [-1,1], , (d)[-i,i]- -, , 2, , 2, , 61. The solution set of the equation, /, , sin, , + cos, , 2, , Then the possible values of 'k' for which g is surjective, function, is, 1, (a), 2, \, 2, , 2 J, , -sin, , X, , (a) [-l,l]-{0}, (b) (0,1] <j{—1}, (c) [-i,0MiJ, (d) [-l,l], , ^/x(x -1) + cosec -1 y 1 + x - x 2, , x, , 2, , 68. Number of values of x satisfying the equation, , cos (3 arc cos (x -1)) = 0 is equal to, (c) 25*, , (d) 40*, , ---- sin “ — = 0 (a, b, 0), then the maximum, a, b, value of b2x2 +a2y2 + 2ab xy sin 0 equals, , 69. Which one of the following function contains only one, integer in its range?, , (b)(a + b)2, , (a) ab, ,2, , [Note sgn (fc) denotes the signum function of k.], , (d) a2b2, , (c)2(a + b):, , f, , 1, , 64. The value of S tan, , (a)f(x) = |cos, , X, , '1 -x2>|, J + XJ, , is equal to, , /2 + 5r+7,, , r=1, , (b) 1, (d) 3, , (a) 0, (c) 2, , i y, , co, , (b) g(x) =sgnl x + — |, \, , (a) tan-13, 1_, , To, , ., = —, IS, , (b) 1, (d) 3, , (a) 0, (c)2, , 62. The value of the angle tan-1 (tan 65° - 2 tan 40°) in, , (c) sin, , 67. Number of values of x satisfying simultaneously, sin-1 x = 2 tan-1 x and, , tan, , degrees is equal to, (a)-20°, (b)20°, , (<*> -1.1], L 2 J, , 1-x2, X, , 63. If cos, , ■, , (c), , X = cot, , X, , 7t, , 1.1 + *',, , J2, 2, , O’-T, , (c)(-i, i)- -, , I, , 66. Let g : R —> I 0, — is defined by g(x) = cos, , xj, , ”>7, , (c) h(x) = sin2 x + 2sinx + 2, , (d) cot-12, , (d) k(x) = cos-1(x2 - 2x + 2), , 70. If range of the function /(x) = tan-1(3x2 + bx + 3), xe R, , 65. The range of the function,, is 0, — |, then square of sum of all possible values of b, , /(x) = tan, , - tan, , 1-x, , x is, will be, (a)0, (c) 72, , (b) {-(rc 14), 3n/4}, (d){3n/4}, , (a) {rt/4}, (c) {k/4, —(3tc/4)}, , (b) 18, (d) None of these, , g Inverse Trigonometric Functions Exercise 2:, More than One Correct Type Questions, i, , 55n, tc i, , i, , 2jt i, , 71. Let 0 = tan-1 tan— and 0 = tan-1 - tan — then, I 44 J, I, 3J, , (a) 0 > 0, , 72. Let/(x) = e cos-1 sin(x + 7t /3) then, Su/18, , 13k/18, , (b) 40 - 30 = 0, , (c)0 + 4> = ^, 12, , (d) None of these, , • (c), , \, , 1=*" '12, 4J, , = elln/12, , www.jeebooks.in

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www.jeebooks.in, ’6)+tan"(9}is, Chap 04 Inverse Trigonometric Functions 365, , 81. If cos-1 x + cos-1 y + cos, , 73. If the numerical value of tan< cos, , z = 7C, then, , (a) x2 + y2 + z2 + 2xyz = 1, , -, HCF(o, b) = l, then, b, (a) a + b = 23, (c) 3b = a + 1, , (b) 2(sin-1 x + sin-1 y + sin-1 z) = cos x + cos" y + cos z, , (c) xy + yz + zx = x + y + z- l, (b)a-b = ll, (d) 2a-3b, , (d) (X+x) + i’' + y) + G + z), , 74. Let /(x) = sin-1 x + cos ~1 x. Then — is equal to, (b) f(k2 - 2k + 3), k, , (a)/, , g, , 82. 2tan(tan-1(x) +tan-1(x3)) where xe R- {-1,1} is equal, to, , R, , (a), (c)/, , 75, cos, , 1, ,ke R, 1 + k2,, , (d)/(-2), , (c) tan(cot"*(- x) - cot-1(x)), (d) tan(2cot"‘x), , , , 2 ^-1, (a) x" =------2, , (b) x2 =, , 5-1, 2, , 83. Let /(x) = sin-1 |sin x| + cos-1 (cos x). Which of the, , 2, , (d) tan(cos“1 x) =, , 5 -1, 2, , 76. The value(s) of x satisfying the equation, sin 11 sin x | = ■Jsin- *| sin x | is/are given by (n is any, , integer), (a) nit -1, (c) nit +1, , xW>, zN>, , following statement(s) is/are TRUE?, (a) /(/(3)) = n, (b) /(x) is periodic with fundamental period 2lt, (c) /(x) is neither even nor odd, (d) Range of /(x) is [0,271 ], 84. If/(x) = sin-1 x. cos-1 x. tan-1 x, , (b) nn, (d) 2nn + 1, , . cot-1 x.sec-1 x. cosec-1x, then which of the, , 77. If (sin-1 x + sin-1 w)(sin-1 y + sin-1 z)= 7t2, then, , D=, , 2x, 1-x2, , (b) tan(2 tan-1 x), , x = tan-1 x, then, , (c) sin(cos-1x) =, , >6, , yN*, (NltN2,NitN4eN), , following statement(s) hold(s) good?, (a) The graph of y = /(x) does not lie above x axis, (b) The non-negative difference between maximum, 3, , minimum value of the function y = /(x) is, , (a) has a maximum value of 2, (b) has a minimum value of 0, (c) 16 different D are possible, (d) has a minimum value of - 2, , 78. Indicate the relation which can hold in their respective, domain for infinite values of x., (a) tan| tan"’x| = | x|, , (b) cot| cot“‘x| = | x|, (c) tan-1] tanx| = | x|, , (d) sin | sin-1 x| = | x|, , 1, , 79. To the equation 22”/cos 'x, , ., , 2JI/C0S-1 X, , -a2 =0, , 2, , has only one real root, then, (a)l<a<3, (b)a£l, (c)a<-3, (d)a£3, , (c) The function y = f(x) is not injective., (d) Number of non-negative integers in the domain of, /(x) is two., /, + 3 tan, and, 85. Let a = 3cos, , lV28>, , P = 4 sin, , - 4 tan, , V 10 7, , then which of the following does not hold(s) good?, (a) a < n but p > it, (b) a > it but p < it, (c) Both a and P are equal, (d) cos(a + P) = 0, , 86. Let function f(x) be defined as, /(x) = |sin-1 x| + cos, , Then which of the, , 80. sin-1 (sin3) + sin-1 (sin 4) + sin-1 (sin5) when simplified, , reduces to, (a) an irrational number, (b) a rational number, (c) an even prime, (d) a negative integer, , following is/are TRUE., (a) /(x) is injective in its domain., (b) /(x) is many-one in its domain., (c) Range of f is singleton set, (d) sgn(/(x)) = 1, where sgn x denotes signum function, of x., , www.jeebooks.in

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www.jeebooks.in, 366, , Textbook of Trigonometry, , 87. Which of the following pairs(s) of function is(are), , 89. Let f: I — {— 1,0,1} —> [— 7t, 7t] be defined as, , identical?, , (a) f(x) = sin(tan“‘ x), g(x) =, , f(x) =2 tan, , X, , 7i + x.2, , (b) /(x) = sgn (cot-1 x), g(x) = sec2 x - tan2 x, where sgnx, denotes signum function of x., x2 -1, , In I cos, , (c) /(x) = e, , , g(x) = COS, , x2 -1, x2 + 1,, , x - tan, , f[1-x2y 2,, , , then which of the, , following statements(s) is (are) correct?, (a) f(x) is bijective, (b) /(x) is injective but not surjective, (c) f(x) is neither injective nor surjective, (d) /(x) is an odd function, [Note : I denotes the set of integers], , (d) /(x) =sin, , 2x, I, g(x) = 2tan x, 1 + x2, , 88. The value of E cot, , —1, 2, 90. If logx = —, logy = - and P = log, , (n2 + n +1) is also equal to, , (sin(arc cos yl - x, , n=1, , /, , f, , (a) cot-1(-l) + sec-l(l) - cosec-1(l), , (c) minimum value of the function f(x) = tan, , 1-x2, 1 + X2, , *y, -4, , (a)P=^-, , 0»p+e=-, , 9, , (d) cos, , , then, , Q = log cos arc tan, , (b) cot~’(2)+ cot-1(3), , -2, , cos41 —, 4., , (C)P-Q = —, , (d)Q="5, , g] Inverse Trigonometric Functions Exercise 3:, Passage Based Questions, Passage I, , 93. The range of /(/i(x)) is, , (Q. Nos. 91 to 93), , Let S denotes the set consisting of four functions and, 5 = {[x], sin x, |x |, {r}} where, {x} denotes fractional part, and [x]denotes greatest integer function. Let A,B,C are, subsets of S., , (b), , (c)f°,7, \, , (d), , 2, , bf], , Passage II, , Suppose, , (Q. Nos. 94 to 96), , A : consists of odd function(s), , B : consists of discontinuous function(s), , and C : consists of non-decreasing function(s) or increasing, function(s)., If /(x)e A n C;g(x)eBn C;h(x)e B but not C and /(x)e, neither A nor B nor C., , Let/be a real-valued function defined on R (the set of real, numbers) such that /(x) = sin~1 (sinx)+cos-1 (cosx), 94. The value of /(10) is equal to, (a) 6n - 20, (c) 20 — 7tc, , (b) 7n - 20, (d) 20-6n, , 95. The area bounded by curve y = /(x) and x-axis from, , Then, answer the following., , 91. The function /(x) is, , < x < 7t is equal to, , (a) periodic, (b) even, (c) odd, (d) neither odd nor even, , (a)^, 4, , (b)^, 2, , (c)7t2, (c)n2, , (_’/', (d)~, 8, , 96. Number of values of x in interval (0, 3) so that /(x) is an, , www.jeebooks.in, 92. The range of g(/(x)) is, (a) {-1,0,1}, (c){0,l|, , (b) {—1,0}, (d) {-2,-1,0,1}, , integer, is equal to, (a) 1, (c)3, , (b)2, (d) 0

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www.jeebooks.in, !, =, , Chap 04 Inverse Trigonometric Functions 367, , Passage III, , i, , l+9x2, , (Q. Nos. 97 to 98), , Consider a real-valued function, /(x) = -(/sin-1 x + 2 + -Jl-sin, , x, , (c)[-l,sinl] (d)[-1,0], , ZI., 101. If(x-l)(x2 +l)>0, then sin - tan-1, , I22, , (c)[1,76], , (d)[A>/6], , Passage IV, , (b)4=, V2, , (c)-l, , (d) None of these, , 2 tan-1 x,|x|< 1, , (Q. Nos. 102 to 104), Forx, y, z,te R, sin-1 x + cos-1 y + sec-1 z> t2 - Jlitt + 3k, , = • -7i + 2tan-1 x,x> 1, , 7t + 2tan-1 x,x< -1, , 102. The value of x + y + z is equal to, , (a) 1, (c)2, , 2 tan-1 x, |x|< 1, 2x, , sin, , = - K-2tan-1 x, x> land, , 1+x2, , 4x, , 99. sin"1, , x = n/2 for-l<x< I, , + 2 tan 1, , —J is independent of x, then, , vx2+4,, , (a)xe[-3, 4], (c) x e [-1,1], , (b) 0, (d)-l, , 103. The principal value of cos " ’(cos 5t2) is, , -(ft+ 2 tan-1 x),x< - 1, , sin' x+cos, , tan 1 x is, , Passage V, , Given that,, , 17?, , 1-x 2, , (a) 1, , (Q. Nos. 99 to 101), , 2x, , 2x, , equal to, , (b)[l,73], , tan, , (d) None of these, , (c)(-oo,m-i), , 98. The range of /(x) is, (a)[0,^], , = - — + 2 tan 13x, then x e, 2, (b) (-1, oo), , 97. The domain of definition of f(x) is, (a)(b)[sinl,l], , 6x, , 100. If cos-1, , (a)7, , (b)-, , (c)I, , W,T, , 104. The value of cos-1(min{x, y, z}) is, , (a)0, , (b) x e [- 2,2], (d) x € [1, «), , (c)n, , <d>7, , g Inverse Trigonometric Functions Exercise 4:, Single Integer Answer Type Questions, (x-2), , ', , 105. Let/(x) = tan, , , then 26f' (1) is, , min(x2 + 4x + 7))}., , is [a, b\ then find the value of —., 7a, , 107. If Z 2 arc cot, n=0, , denotes greatest integer function), , 13, 110. If sin (30° + arc tanx) = — and0< x< 1, the value of x is, fl v 3, , n2 +n + 4A, , where a and b are positive integers with no, , = to, then find the value of, , 2, , 7, , common factors. Find the value of, , k., 108. Find the number of solutions of the equation, ( 5, , >, , is equal to (where [.], , z, , vx2 + 2x + 2>, , 106. Let f(x) = (arc tan x)3 + (arc cot x)3. If the range of f(x), , sin- z, , 111. Let f: R-> ^0,, , a + h'l, 2 J‘, , defined as /(x) = cot-1 (x2 -4x + a)., , 5x + 6, , tan E cot 1(2r2) =, =l, J, 6x + 5, , Find the smallest integral value of a such that /(x) is, into function., , www.jeebooks.in, 109. lim [{max (sin, , x + cos"1 x)2,

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www.jeebooks.in, 368, , Textbook of Trigonometry, , 112. Let L denotes the number of subjective functions, f : A—> B, where set A contains 4 elements and set B, contains 3 elements, M denotes number of elements in, the range of the function, /(x) = sec-1 (sgnx) + cosec-1 (sgn x), where sgnx, denotes signum function of x and N denotes coefficient, oft5 in(l + t2)5 (1 + t3)8., , 115. Let /(x) = cos(tan-1 (sin(cot-1 x))). The simplest form of, , /(x) can be written as, , f x2 +^Y/2 . Then the value of, , (A + B)is, , 116. Find the value of A, for which, cos 2a sec 20 + cos 20 sec 2a, tan-1, , X, , Find the value of (N - LM), , 113. Number of solution(s) of the equations cos-1 (cosx) = x'2, is, 114. If cos -1 (x) + cos"1 (y) + cos -1 (z) = it(sec2 (u), , + sec4(v) + sec6(w)), where u, v, w are least, non-negative angles such that u < v < w, then the value, of x2K” + y2002 + zlm + 3671, is, U+V+w, , = tan-1 {tan2 a + 0) tan2(a - 0)} + tan-1 1, , 117. The least value of n for which, (n -2)x2 4- 8x + n + 4 > sin-1 (sin!2) + cos 1 (cos 12),, Vx g R, where n e N, is, , I, , 118. IfOccos-1 x< 1 and 1 +sin(cos-1 x) + sin2(cos-1 x) +, , sin3(cos-1 x) +... oo = 2, then the value ofl2x2 is, 119. The number of real solutions of the equation, ^/1 + cos2x = ^2 sin-1 (sinx), - it < x < it, is, , g Inverse Trigonometric Functions Exercise 5:, Statement I and II Type Questions, ■ This section contain 6 questions. Each question contains, Statement I (Assertion) and Statement II (Reason)., Each question has 4 choices (a), (b), (c) and (d) out of, which only one is correct. The choices are, (a) Statement I is True; Statement II is True; Statement II is a, correct explanation for Statement I;, (b) Statement I is True; Statement 11 is True; Statement II is, NOT a correct explanation for Statement I;, (c) Statement I is True; Statement II is False;, (d) Statement I is False; Statement II is True., , 120. Statement I y = tan-1 (tanx) and y = cos-1 (cos x) does, , .,, , f71 31tA, , ,, , 2it-x,, , y = cos 1(cos x), x,, 121. Statement I sin 1, , 31t, , xe K,---, , 2 _, It, xe —, It, .2, , > tan -i, , Statement II sin-1 x > tan"1 y for x > y, V x, y e (0,1), , \2 -Jz), , 11, >sec 1 -1 + -F, , 2 JI), , Statement II cosec-1x>sec-1 x, if 1 < x < ^2, , 123. Let/(x) = sin 1, , 2x, , J + x2,, 2, , Statement I f' (2) = — and, 5, , Statement II sin 1, , 2x, , = it -2 tan-1 x,Vx>l, , J + X2;, , not have any solution, if xe —, —, \2 2 J, , it 3n, Statement II y = tan 1 (tan x) = x - it, x g, and, 2* 2, , -1+*, , 122. Statement I cosec, , 124. Statement I sin 1 2x + sin-1 3x = —, 3, , =>x = J— only., V 76, , and, , Statement II Sum of two negative angles cannot be, positive., 125. Statement I Number of roots of the equation cot-1 x, cos-1 2x + it = 0 is zero., , Statement II Range of cot"1 x and cos -1 x is (0,it) and, [0, it], respectively., , www.jeebooks.in

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 369, , g Inverse Trigonometric Functions Exercise 6:, Matching Type Questions, , 126. Let fj = (sin 1 x)isin, : ’lx,t2=(sin-1x)C0S‘Ix,, , 127. Match the following items of Column I with, Column II, , = (cos’1x),u, sin*’,x, t3 =(cos"*, 'x,t4 =(cos-1 x)cos’‘x,, , Match the following items of Column I with Column II, , Column 1, A. x e (0, cos 1), B., x e cos 1,-?=, \ V2., C., • .1, x e fl, -7=, sin, , (P), (fl), , Column II, h >t2>t^>t3, <( > (j > #] > t2, , (r), , r2 >, , Column II, , Column I, A. sin-*x + x > 0, for, , (pT"7<0, , B. cos-1 x - x > 0, for, , 7q) x 6 (0,1], , C. tan-1x +x < 0, for, , F)~x7[-7,6j', , D. cot-Ix + x>Q for, , (s), , x>0, , > t3, , 172, , D. x e (sin 1,1), , (s)_ _ t3> /4 > /] > t2, , Inverse Trigonometric Functions Exercise 7:, Subjective Type Questions, 10, , 128. Solve sin-1 — J + sin 1, x), 129. Solve tan-1, , x+r + tan, x-ly, , r = lJ =1, , 1|, | = tan-1 (-7), { x J, , 134. Find the value, n, , lim £cos, 130. If a, b, c are positive, show that, tan-1, , a(a + b + c), , be, , + tan-1, , cos2 0, c(a + b + c), = 71, ab, , cosec “1710 + cosec "15^0 + cosec “17170, , +... + cosec -1 -J(n2 + l)(n2 +2n + 2), , + 7sin-1, , 28 then find the maximum, , cos 1 x2 + sin 1 x2, x3, , -1)^ + 1)^ + 2), , k(k + l), , \, 7, , ntan0, , b(a + b + c), ac, , 131. Find the sum to the n term of the series, , Jsin-1 *1, , +, , -i, , \, , + tan 1, , 132. If Xj e [0,1] Vi = 1,2,3, value of, , 10, , 133. Find the value of y y tan, , cos 1 X3, , x, cos 1 x4 +... + ^/sin -1*28, cos 1 *1, , cos2 (a -0), , y, , Prove that: 0 = - a - tan, 2, , n-m, tana, n+m, 7., , 136. Prove that:, _lz je. n 7t K i T, fit 0"), tan (e ) = — +------ In tad------- L where 'n is an, 2 4 2, \4 2), integer., , 137. If the quadratic equation;, .2 +2x+^p2 -p + -) = 0have meal roots, then, find all the possible value of cos a + cos 1 {1, , www.jeebooks.in

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www.jeebooks.in, 370, , Textbook of Trigonometry, , 0 Inverse Trigonometric Functions Exercise 8:, Questions Asked in Previous 10 Years Exam, -I r, , i, , rwt, , ■ r, , , w. ■■ RB ■, , bi", , a vw nv xw im^w w« a«, , ■ B rr. i~i ! Lin rirnu, , (i) JEE Advanced & IIT-JEE, , /, , 23, , 138. Ifa = 3sin, , and P = 3 cos, , — , where the, l \ 23, , inverse trigonometric functions take only the principal, values, then the correct option(s) is/are, [More than one correct option, JEE 2015 Adv. ], (a) cos P > 0, (b) sin p < 0, (c) cos (a + p) > 0, (d) cos a < 0, , 139. If 0 < x < 1, then, , + x2 [{x cos (cot, , x), , ZkS, , (a) —, 25, , 25, , (b) —, 23, , <c)s, , 142. If x, y and z are in AP and, , (c,u, , (d)TiT^.2, , Then, the value of y is, 3x — x2, 1 -3x2, (C), , 2x, , I, , ,, , ,, , 1, , , where | x | < —=., kl-X2;, V3, , [2015 JEE Main], 3x + x2, (b)^4, 1 -3x2, , (d)^-A-, , |, |, k4/, , + cosec, , (a) 1, (c) 4, , 145. If cos, , x-cos, , , then the value of x is, [2007 AIEEE], , -i y, , ?, — =a,then4x - 4xy cos a + y2 is, 2, [2005 AIEEE], , (b) 4sin2 a, , (c) 4, , 1 + 3xz, , 2, , (b) 3, (d) 5, , equal to, (a) -4 sin2 a, , 3x + x2, , 3x — x2, 1 +3x2, , [AIEEE 2008], , <d), (d) 77, , 144. If sin, , (ii) JEE Main/AIEEE, x + tan, , 5 ■ r1 - + tan, is, 3 ', 3, (b)^, (b) 77, , yi + x2, , y = tan, , x, tan-1 y and tan-1 z, , (b) 2x = 3y = 6z, (d) 6x = 4y = 3z, , (b)x, , 140. If tan, , 23, , tan -1, , [2013 JEE Main], , are also in AP, then, (a) x=y=z, (c) 6x = 3y = 2z, , [Single correct option, IIT-JEE 2008 3M], , (c) X-Jl + X2, , [2013 JEE Main], , a \23, , 143. The value of cot cosec, , + sin (cot-1 x)}2 -1]1/2 is equal to, , J, , k=1, , \, , n=l, , W, , n, , 1 + y 2k > is, , 141. The value of cot< y cot, , (d) 2sin2a, , Answers, Exercise for Session 1, 1.1/2, , 2.-1, , 4. R, , 5. (-~, - 1] u[l, «), , 5. 0, , 10, , 3. (-«>,-3] u[-2,-1] u[0,«), , 1. n, , 2. 2, , —n, 3.^, 10, , 1.1, 5, , 10.x=i, 2, , i K =~---19 ~5. x = J5 + 4-/2, 3., 9, 9, , 6.x=l, 2, , 1, “, 2.—, 16, , 2., 9, 3.—, 25, , 2V5, , 4.—, , 5, , Exercise for Session 7, , Exercise for Session 3, 2. not defined, , 9. x = 1, , Exercise for Session 6, , 4. 13- 4tc, sin-,(sin0) = 3n-0;, cos '(cos0) = 0 - 2jc, 5., tan-1 (tan 0) = 0-3n; cot 1 (cot0) = 0- 2it, , 1.^, 2, , 7.", 2, , Exercise for Session 5, , Exercise for Session 2, ** 2n, 5, , , —, 3n, o., , 3.^, 36, , Exercise for Session 4, , 4. 11, , 5. x= ±1, , 3 sin- x, , 1. j=sin’'(3x-4?) = < 7t — 3 sin 1 x, , ;, , ;, , 1, 1, 2, 2, 1<x<1, 2, , —Sx<-, , -n- 3sin- x ; -15x<--, , www.jeebooks.in, 1.2E, 8, , 2., , 1 _ nA, 4 2j, , 3.1, 2, , 2

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , Graph of y= sin’*(3x- 4X3), , 6., , 371, , Ya, , Y, , k/2, , n, 2, , 0, , V3, , 41, , *X, oo, , r-n/2, n, 2, , 7., , Yf, , n, , 2. y = tan, , <3x-?>, J-3X2,, , 1, , -~=<X<-=, , n + 3 tan 1 x, , -OO < X< --4=, , V3, , J3, , --n/2, , V3, 1, , -n+3tan’*x ;, , Graph of y = tan, , 1, , 3 tan’ x, , *X, , 0, , -7= < X< oo, , V3, , 'ix-x*', , Chapter Exercises, y, n, 2, , I- (b), 7. (a), , 13. (b), , 19. (b), 25. (d), 31. (c), , n, 2, , 3., , K, , jc, , 4.(d), , 5-(a), , 8- (b), , 9. (a), , 10. (b), , 11. (d), , 6. (b), 12. (b), , 14- (a), 20. (b), , 15. (a), , 16. (b), , 17. (d), , 18. (b), , 21-(a), , 22. (d), , 23. (d), , 24. (d), , 26. (b), , 27. (d), , 28. (a), , 29. (a), , 30. (c), , 32- (b), , 33. (a), , 34. (b), , 35. (c), , 36. (d), , 41.(d), , 42. (c), , 37. (b), , 38. (a), , 39. (c), , 40. (d), , 43. (d), , 44. (b), , 45. (c), , 46. (c), , 47. (d), , 48. (a), , 52.(c), , 53. (a), , 54. (a), , 58. (b), , 59. (a), , 60. (c), , 66. (c), , 49. (d), , 50. (a), , 55. (c), , 56. (c), , 51. (c), 57. (b), , 61. (c), , 62. (c), , 63. (d), , 64.(c), , 65. (c), , 67. (c), , 68. (d), , 69. (d), , 70. (a), , 71. (b,c) 72. (b,c), , 79. (b,c), , 75. (a,c), 80. (b,d), , 76. (a,b,c) 77. (a,c,d), 81. (a,b), , 83. (a,b), , 84. (a,b), , 87. (a.c), , 88. (a,b,d) 89. (c.d) 90. (b,c,d), , 82. (a,b,c), , 92. (d), , 93. (b), , 94. (b), , 95. (b), , 96. (c), , 97. (c), , 98. (d), , 99. (b), , 100. (a), , 101. (c), , 102. (d), , 103. (b), , 104. (c), , 105. (9), , 106. (4), , 107.(1), , 108. (0), 114.(9), , 120. (a), , 91- (b), , Ya, , n, 2, , V2, , >X, , ■4—, , 1, , <2, , ., , 85. (a,b,d)86. (a,d), , >X, , 1, , 0, , 0, , 3.(c), , 73. (a,b,c)74. (a,c), 78. (a,b,c,d), , -n/2, , 4., , 2. (a), , 109.(3), , 110.(8), , Hl-(4), , 112.(4), , 113.(3), , 115.(3), , 116.(2), , 118. (9), , 119. (2), , 121. (a), , 122. (a), , H7.(5), 123. (a), , 124. (a), , 125. (a), , 126. A -> q; B -> s; C -> r, D -> p;, , 127. A -» q; B -> r, C -> p j; D -> q,r,s;, 129. No Solution, , 128. x = 13, , n, 2, , i, , 131. tan’(n+1)-— 132.7n, ' 4, , 5., , 133. 25 k, , 134.6, , n+1, :l, 137. —-land —, 3, 3, , Y, n/2, , 138. (b,c,d), , 139. (c), , 140. (a), , 143. (b), , 144. (b), , 145. (b), , 141. (b), , 142. (a), , www.jeebooks.in, 0, , -n/2, , -►X, oo

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www.jeebooks.in, 374, , Textbook of Trigonometry, , /, , 23. [cot - ’ x] + 2[tan-1 x] = 0, , =>, , =>, , [cot - ’ x] = 0, [tan-1 x] = 0, , or, , [cot-1 x] = 2, [tan-1 x] = - 1, , Now,, , [cot- ’ x] = 0 => x e (cot 1, “), , [cot-1 x] = [tan-1 x] = 0, x e (cotl, tanl), , x=0, , =>, , 29. Let /(x) = x3 + bx2 + ex + 1, /(0) = 1 > 0, f(- 1) = b - c < 0, So, a e (-1,0), So, 2tan-1(cosec a) + tan-1(2sinasec2 a), , ( 3k, Now, cos-1 x = cos *(sin0) = cos 1 — cos------0, , = 2tan-1f——, , I2, , \sina., , = it - cos-1fcosf— - 0 11, , I 2 • )), , 2 sin a, , + tan-1, , <1 -sin2a, , = 2 tan-1[ —— | + tan ’(sina), k sina ), , ( 3tc, a_ . 337t, tc A, = n------- 0 , as 0 <------ 0 < 7t, , ), , V2, , = 3 tan-1 x tan-12x tan-13x, , 24. Let sin-1 x = 0 => x = sin0, — < 0 < —, 2, 2, , I2, , 1, = = 1 => a = —7=, , 28. Expanding, we have, (tan-1x)3 + (tan-l2x)3 + (tan-,3x)3, , [tan-1 x] = -1 => x e [- tanl, 0), , k, , n, 4, , 2, , Jl - a 2, , [cot-1 x] = 2=>xe (cot3, cot2], Hence, no such x exists., Thus, the solution set is (cotl, tanl), , x, , a, , a, , =>, , [tan-1 x] = 0=>xe(0, tanl), Therefore,, , tan-1, , 2, , =2, , (as sina < 0), , - 7C, , It, , = 0- —=sin-1 x---2, 2, , Hence, sin- ’ x + cos-1 x = 2 sin-1, , 30. Given, u = cot-1Vcos20 - tan-1 Vcos20,, , 71, X------, , Put, , 2, , cos20 = tan20, u = cot-1(tan0) - tan-1(tan0), , 25. asin ’x-hcos *x = c, hrr, , = --0-0 = --20, 2, 2, , We have, frsin- ’ x + b cos-1 x = —, 2, hrr, , I 71*, , fih., , X'+, , 26. sin-11 x-2| + COS ’(1 -|3-x|) = y, , 0, , V, , |x-2| = l -|3-x|, |x-2| + |3 - x| = 1, |x-2| + |3-x| = |(x-2) + (3-x)|, (x-2)(3-x)£0, 2^x<3, , or, or, or, , 2, , 31., , bit + 2c, sin-1x =, or, a+b, %a + b), bit, +, 2c, _ ita - 2c, -i, n, COS X = —, 2 2(a + b) 2(a + b), itab + c(a - b), => asin-1 x + hcos-1 x =, a+b, , or, , I, , sinu = sin---- 20 = cos 20 = tan 0, \2, ), , Adding (a + b) sin-1 x = — + c, , 27. According to the question,, 2, 3, sin-1 —7= + sin-1 -7== + sin-1 a - it, 45, V10, X, (, a, = 7t, tan-12 + tan-‘3 + tan-1, , Thus, /(x) is not differentiable at x = 0, , K, , 32., , X'*, , 0, x=-1, , X=1, , www.jeebooks.in, =>, , it + tan, , -1, , 2+3, ----------- + tan, l-(2)(3), , a, , 2, , = 7C, , From the -graph, it is clear that, f(x) is not differentiable at, x = 1, -1

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 375, , 33. Given, f(x) = tan”‘(x2 - 18x + a) > 0, =>, , = sini -ify-i, —, , I 2 ,, , tan”’(x2 — 18x + a) > 0, , x2 - 18x + a > 0 => 1822 - 4a < 0, , =>, , => ex sin” ip"1, , -sin, , I 2, , 182 18X18, =>, a>— =---- =8i, 4, 4, =>, a >81 => a 6 (81, «>), 34. /(x) = sin”12x + cos’12x + sec’12x, , I 2, , sin -if y -11, , I 2 J, , =>, , 1 - sin, , I 2 J, , /, , /(x) = — + sec-I2x, 2, Graph of sec"12x is as following, , sin, x = log, , -ifl —, 2 J1, , 1 - sin, , k 2, , ■ -il x~ 1 I, , sin, , 7C, , ------, , I 2 /, , r‘(x)=iog, , ./x-1, A — 1, , . -1, , 1 - sin, 7C, , 2, 1, 2, , 5, , >X, , = cos”’(cos(2(67-50))), = cos-,(cos(1350)), , =7+ 0 = 7, , 4, , (/wu., , 40. /(x) = tan-1, 3n, , „, , Sum = — + — = 2n, 2, 2, , >b, , ----- +, , 35. Given, tan”, , i c+a, , c, a+b, c, , (712-2)x2, x* + 2x2 + 3, , ', , 2(75 -1), , = tan"1, , x2 4- \ + 2, x, J, , As x2 + — > 275 [using AM > GM], x, , 7C, , x,22 + —r + 2 £ 2 + 275, x2, , 4, , = tan-1, , ab + b2 + c2 + ac, , =>, , f, , \, , c+a a+b, , ., , | ——, 2, , 39. cos” '(cos(2 cot” '(75 -1))), , 0, , o, , v, x, , ---------- ------------ 2-------------------- = 1, , ac + be + a + ab -be, , ab + b2 + c2,+ ac = ac + a2 + ab, , 41. tan”, , , <J1 + x2 - 1, , 2(73-1), 2(75 + 1), , It, , 12, (x*0), , = 4°, , X, , b + c =a, , ■jl + X.22 -1, , z.AABC is right angled at A, , =>, , 36. By graph, clearly it has 5 real roots, y=sinx, ., sin”1 (sin x), , 3n/2, , = tan 4°, , X, , =>, , =>, =$>, , 0, , 71 + x2 = 1 + xtan4°, , 1 + x2 =2xtan4° + 1 + x2tan24°, 2 tan 4°, x = 0 or x = -------;— = tan 8°, 1 - tan2 4°, , Since x * 0, we have x = tan8°, 2, , 37. -1, , 42. From the given equation sin29 - 2sin9 + 3=5*“ y + 1,, , x < l,y > 0, , Both will be equated for infinite values of x and y. Therefore, infinite many solutions., , 38. y = 1 + 2sin, , ,x£0, , we get, , (sin9 - I)2 + 2 = 5*"’y +1, , LHS<6,RHS>6, Possible solution is sin9 = -1 when LH.S. = R.H.S., =>, cos29 = 0, , www.jeebooks.in, Z+b, , =>, , y -1 = 2sin, , ' ex ', , y+ 1,, , =>, , cos29 -sin 6 = 1

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www.jeebooks.in, 376, , Textbook of Trigonometry, , 43. y = |x2 - 1| =| tan-1| x||, , 48., , n/2, , 2e + 4, , 3, , 2e + 4, , > e => Lnl, , 3, , . 5071, 5071^, , 49. sin, , . _j . (1471, , . -j. <5071^1, 5O7t, , - sin — | = -sin sin | — | = - sin sinl —1, 9, 9, 47T1, , = -sin sin 27t------k, . 9 J, , y=|x2-i|, , y=|tan-’|x||, , = - sin', , From graph, clearly it has 4 solutions., , 31n, , cos cos, , 44. 1 + tan2(tan-1 x) ~(sec2(sec-1 x) -1), , 9, , _i, f317t^, = cos cos, 9, , f., , = 1 + (tan(tan-1 x))2 - (sec(sec-1 x))2 + 1, , 571, , 5tc 5rt, = cos cos— =, 9, 9, , = cos cos 47t-----\, 9 ., , = 1 + x2 - x2 + 1 = 2, , „, (47C, 5tt, Hence, sec----- 1- — = sec n = -1, , 45. The equation 3x2 + 6x + a = 0 must have equal roots, , I 9, , 9., , So, D = 0, 36 —12a = 0 => a =3, , 46. tan', , 1, . ., -7= + sin, V2, , 1, , -7= - COS, , 50. tan- arc tan(2) + arc tan(20fc) • = k;, , 1, , -7, , V:10, , v5, 1, , = tan, , -7= + tani, V2, 1, , = tan, , =>, , -1, , -i, , 1 - tanA tanB, , — tan 3, 2, , tan 3 - tan', , 2, , 3-tan' ___ 2 = tan, , 2, , I+’, , tan, , 1 - tan', , 40fc2+19fc + 2 = 0, , or, , 1, , -1, , -?=- tan !1, V2, , 51., , E tan', r=2, , (r—2)-(r—3), , T3 = tan’l - tan-10, , T4 = tan-12 - tan, , = tan, , Tn = tan ’(n - 2) - tan-1 (n - 3), , •J2 + 1, , Sn = tan-1(n - 2) + —, 4, , cot, , _, , = - it + cot, , 4, , _ it, , 0, cos— =, 2, , 3, , + tan, , 1 -4, , 4 „, - = 2 tan, 3, , 4, , 4, 4, - => tan0 =, 3*, 3, , Now,, , sec a = sec 20 =, , 4, , 4, , 53. tan, , 1 + COS0, , 3, , 2, , 4, , ^tan2Aj + tan-1 (cot A) + tan-1 (cot3 A), , 3, , 2x, (using 2 tan-1 x = tan' l-x2), Let 0 = tan, , it _ 3tc, , 0 ; where cos0 =-1, 52. x = -ltl4;y = cos—, 2, 8, and, , 1, , 1, , V2 -1, , 2, , 1, 47. a = 2 tan' - + tan, 2, , r=2, , T2 = tan-1 0 - tan-1(- 1), , Now,, , =>, , n - cot, , = E (tan 1(r-2)-tan-1(r-3)), , l + (r-3)(r-2), , 1, , = - tan', , = - cot, , 19, 40, , 2, , tan, , + k2 = -, , Now, sum of solutions,, , 2J, , 1, , 2 + 20k, = k=> --------------- = k, 1 - (2)(20k), , 1, 2, , i, , tan, , tanA + tanB, , -i 1, , = tan, , -tan2A | + tan', .2, J, , cot A + cot3 A, 1 - cot4 A, , + it, , [ 0 < A < — => cot A > 1, V, 4, , 1 + tan20, , 1 -tan20, , 1 + (16/9) 25, 1-(16/9) “-7, , cot A(1 + cot2 A), tan A, + it + tan', 1 - tan2 A, (1 -cotzA)(l + cot2 A), /, tanA, cot A, = 71 + tan, + tan', J - tan2 A,, 1 - cot2A, , = tan, , www.jeebooks.in, 25, Hence, sec 0 =-----7, , = it = 4tan-1(l)

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www.jeebooks.in, Textbook of Trigonometry, , 378, , On squaring both sides, we get, x2y2, . 2n 2xy, , x2, , a 2b 2, , a2, , 68. Let 0 = arc cos(x - 1), , ab, , x2y2, , y2, , b2, , a2b2, , b2x2 + a2y2 + 2ab xysin0 = a2b2 cos20 < a2b2, , =>, , _______ 1_______ ', , 64. an = tan, , cos30 = 4cos30 - 3 cos0, , Now,, , So,, , 4y3 - 3y = 0, where y - x - 1, , = tan, , l+(n+3)(n + 2), , 1 + (n + 3)(n + 2),, , Hence, three values of x., , = tan"’(n + 3) - tan”‘(n + 2), , Aliter:, cos (3 cos"1 (x - 1)) = 0, , S„ = /(n) - /(I) = tan"1 (n + 3) - tan-13, , S„ = — - tan"13 = cot"13 = sin"1 -i=, 3 = sin, ”2, V'10, , 65., , tan, , cos-1(x - 1) = (2n + 1) —, neZ, 6, n it 5lt, cos (x-l) = —, —,—, 6 2 6, , ^3-^3, X - 1 = —, 0,-----2, 2, , -it + tan-1(l)+ tan”1 x, if x>l, , 1-x J, , Now, f(x) = ■, , => 3cos 1 (x - 1) = (2n + 1) —, ne Z, '2, , tan'’(l)+tan”1 x, if x<l, , A, , tan-1(l), if x<l, -it + tan”’(l), if x>l, , =>, , /. Range/ = j^, , V3, V3, x = l + —,1,1-----2, 2, , x2-k, 1, (fc+1), 66. We have - <, < 1, V xeR, <1 =>- <1 x2 + 1, 2, 2 1 + x2, k + 1 > 0 and, , 1, 69. (a)/(x) = -cos, 2, , k+ 1 <1, x2 + 1 ~ 2, , r/=[°4), (b) £(x)=sgn (x-b-l, , 4, , \, , "2, , x/, , = (- °o, 0) u (0, <x>), ^={-1.1}, (c) /i(x) = sin2 x + 2sinx + 2, , Hence, k e I -1, —, , 2, , Dh=R, Also, h(x) =(sinx + I)2 + 1, , 67. Domain is x e [-1,1], Given, sin-1 x = 2 tan', , tan, , < it, , <1 +, , k<-2, , =>, , J + X2, , As, 0 £ cos', , => x2-(2k+l)>0, VxeR=>4(21t + l)<0, , I, , "1 -X2, , Df=R, fl-x2>, , k > -1 and x2 + 1 > 2k + 2, , So,, , 73, , 73, , y = ±—,0 => x = l±—,1, 2, 2, , (n + 3)-(n+2), , X, , x, , .•A7**,, , ^=[1.5], (d) k(x) = cos-1(x2 - 2x + 2) = cos"1 ((x - I)2 + 1), , 2x, , = tan, , l-x2?, , ^={1}, , ^={0}, , X, , 2x, , /i-xz, , 1 - x2, , 70. /.0 < tan"*(3x2 + bx + 3) < —, 2, , x = 0or(l -x2)2 = 4(1 -x2), , =>, , Thus, range of 3x2 + bx + 3 is [0,«), , — (0, , x = - 1, 0,1, , tan"1 ^]x(x -1) + cosec, , ■Jl + x - x2, , it, , =±6, , /.Square of sum of values of b=0, , x(x-l)£0nx-x2>0=>x(x-l) = 0, , Now, Eqs. (i) n (ii) gives x = 0,1, , Now, D = b2-4-3-3 = 0 => b2 =36 =>, , Sum of values of b = 0, , 2, , x = 0,1, , 0 5 3x2 + bx + 3 <, , =>, , (1 -x2)(3 + x2) = 0, , -(ii), , 2lt, 71. 0 = tan”1 (tan— | and 0 = tan'-i, - tan—, 3 ., \, 4 J, , www.jeebooks.in, Hence, number of common solution are 2., , = tan, , n, -itan I it + —, K and <|> = tan 1 - tan i it----I 3., \, 4.

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions 379, , 76. The solution of y = Jy is y = 0 or y = 1, , = tan" ’tan — and d> = tan"‘tan —, 4, 3, It, , J A, , If sin- ’| sinx| = l=>x = lorx = it-l, , I1, , = — and d> = —, 4, 3, , But y = sin" ’| sin x | is periodic with period it, so x = mt + lor, mt -1., Again if sin" *| sinx) = 0, x = mt, , 40 - 3<|> = 0 and 0 + <b = —, 12, . (8k kA, sin — + —, , COS, , 7Z, , I 9, , =e, , 3J =e, , cos-’sml^, , 77. (sin"1 x + sin"1 w) (sin"1 y + sin"1 z) = it2, , ’, , .'. sin"1 x + sin"1 w = sin"1 y + sin"1 z = n, , =, , 13k, cos-1, C°’18 =C,13, 1 k Z18, e, , and/- —, =e, \ 4,, , or sin"’x + sin"’w =sin"’y + sin"’z = - n, x=y = z = w = lorx = y= z= w = -l, , --44^7), (-hkA, cos sin -----, , -i, , I *2 J, , =e, cos-1, , =e, , “Gt)., , minimum value, , eK/12, , tana =, , andtanlcos ’-4-tan"1k, 5, 3,, , 2, , -1, , 75. Given, cos"1 x = tan" ’ x, x = cos0 = tan0, cos20=sin0, , . - 1 ± 7T+4, , sm0 =------- ------2, 75-1, sinQ =, 2, x2 = cos2 9 =, , 1, , = 2 and, , '5-1, 2, , Option (a) is correct., sin(cos"1 x) = sin6 = - (75 -1), 2, Also, option (c) is correct, , and, , tan"‘x,, , ifx>0, , -tan"’x, ifx<0, , tan| tan-1x| = tan tan" ‘| x| = | x|, , |cot"lx| = cot"lx; VxgE, , Also,, , cot) C0t"*x| = X, V X 6 R, , x,, tan ’| tanx| =, , (given), sin|sin"’x| =, , if 0 < x < —, 2, , -x, if---- <x<0, 2, X, 6, [0,, 1], x,, -X,, , x G [- 1, 0), , oo =>2 <, <----- —— < oo, 79. 1£ -------- — < «, cos x, 2™ X, , Hence, 2 should lie between or on the roots of, a + -/ - a2 = 0, where t = 2*/co’, t2, 2), /(2)£0 => az + 2a-3>0, , =>, => a g (-, , => sin20 + sin9 -1 = 0, , =>, , -1, , | tan"’x| = tan“*| x|V x g R, , 2, tana + _______ 3_, ,_______ 2, 1 - tana - 3, , 74. sin"1 x + cos"1 x = y.Vx6[-l,l], , =>, , 1, 1, , -1 1, , 78. Since, | tan 1 x | =, , 3, 4, , 3 2, _ 4 + 3 =17_ a, 1-2.3 6, b, 3 4, so,, a = 17,b=6,a + b=23, a - b = 11 and 3b = a + 1, , =>, =>, , zN’, , = - 2 Also, there are 16 different, 1 1, determinants as each place value is either 1 or - 1., , i[4, 73. Let cos" -1 ] = a, that is cosa =4 so that, .57, 5, , 5, 4, , yN', , Hence, the maximum value of, , - 3] U (1, °o), , 80. E = K-3 + K -4 + 5-2% =-2, 81. cos x + cos" y + cos" z = it, , =>, , Also,, , Jt, , sin"’x+sin ly + sin z = —, 2, cos"*x+cos~*y+ cos"‘z = K, cos, , =>, =>, , *y, , x + cos"*y = cos"'(-z), , -^x2 ^-y: = — z, , x2 + y2 + z2 + 2xyz = 1, , Option (d) can be true only if x.y.z > 0; for (c) put, x = y =z = 1/2, , www.jeebooks.in, and, , tan(cos"1 x) = tan0 * j(7s -1), , 82. Let tan"1 x = a and tan"1 x3 = 0, , .'.Option (d) is not correct., , tana = x and tan0 = x3

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www.jeebooks.in, 380, , Textbook of Trigonometry, , „, ., o. 2(tana + tanB), .-. 2tan(a + P)=--------------- =, 1 - tana tanp, , Also,, , 1-x4, , 2x, 1-x2, , 85. a =3 tan, , 2 tana, = tan2a = tan(2tan-1 x), 1 - tan2a, , 2x, 1 -x2, ■ I, , x + x3, , 2, , f K, , = tan 2---- cot, , _i, , I <2, , x, , V34-___, V3, --tan' _5------- 2_ = 3 tan, 10, , P = 4 tan 17 - tan', , 7t, , 0£x£ —, 2, n, 3h, n, 83. /(*) =, 2, 2, 3n, 4it - 2x ; — < x £ 2n, 2, , = 4 tan', , Clearly, f(x) is periodic function with period 2k. The graph of, /(x) is shown below., /-------------- \, x=0, , \, -4—, , n, 2, , 3n, 2, , 86., , 5k, 2, , -+7k, 2, , 7— —, , 3, , = 4 tan' ____ 4_, , 4, , l+“, 4J, , g)->, , f(x) = | sin 1 x| + cos, , x), , Domain of /(x) is {-1,1}, , /(i) = TA1) = v, z, z, , /-------------- x, , 2k/, , = it, , 7 ,, , = tan(cot-1(- x) - cot-1(x)), 2x, , 7VT, , 1-1, , = tan(n - cot-1 x -cot 1 x), , _, , 73, =3, 2, , + 3 tan, , So, function /(x) is injective ., sgn(/(x) = 1 (/(x) > 0), .. . fit 3itl, Range of, , /W“17T, , 84., , Domain of sin 1 x and cos-1 x, each is [-1,1] and that of sec 1 x, and cosec-1x, each is (-«>,-1] u [1, »>), .'.Domain of J(x) must be {-1,1}, .'.Range of /(x) will be {/(-I),/(I)}, where, /(-l)=sin-1(-l). cos-1(-l). tan-1(- l)cot-1(-l), , 87, (a) /(x) = sin(tan-1 x), , Put tan-1 x = 0 => x = tan 9, , li+7, x, , • sec-1(- 1) • cosec-1(-l), 0, , 2, 1, , and, , => /(x) =, , f(l) = 0 {as cos-11 = 0}, , (i) Thus, the graph of f(x) is a two point graph which doesn’t, lies above X-axis., = 0and/(x)min =, ., , 7i+x, , = g(x) => identical functions, , (b) f(x) = sgn (cot-1 x) = 1, V e R, , But g(x) = sec2 x - tan-1 x^ 1, V xeR. (Think ? Domain of, g(x)), . . . x2 -1, x2 + 1 - 2, 2, (c) As —----= 1-----.2^€(-1,1), X2 +. 1, x2 + 1, x2 + 1, , —3 it6, 64, , Hence, |/(x).'max ~ f(xjminl ~, , X, , 31t6, 64, , y., , So, cos', , x2-l, e(0,n], X2 + 1, , So, /(x) and g(x) are identical functions., , ——, (1.0), , -3n6, ’ 64, , (d)/(x)=sin, , 2x, 1 + x2, , - (it + 2 tan-1 x) ,, 2tan-1x, , it -2tan-1x, , x £ -1, , , -1<x<1, , ,, , x£l, , So, /(x) and g(x) are not identical functions,, 88. Tn = cot ’(n2 + n + 1)= tan', , (n + 1) - n, , 1 + n(n + 1)J, , = (tan-1(n + 1) - tan-1n), , www.jeebooks.in, (iii) /(x) is one-one. Hence, injective., (iv) Domain is {-1,1}, .'.Number of non-negative integers in the domain of f(x), is one., , n, , So, E Tn - (tan-12 - tan-11) +, n =1, , (tan-13 - tan-12) +, , + (tan-1 (n + 1) - tan 1 n)

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , = tan-1(n + l)-tan-1l, , 94. We have, /(10) = sin-1 (sin 10) + cos, , n + 1-1, --------------- = tan, _l + (n + l)-l, , = tan, , n, , 381, , (cos10), , = (3rc -10) + (4n -10) = (7tc - 20), , n+2, TC, , =>, , 95. Clearly, f(x) = (n - x) + x = it, V x e —, it, .2' ., , S Tn = tan-11 = —, n=1, , 4, ,, , (a) cot-1(-l) + sec, , 37t, , -1,, , Y., , . k, , K, , „, , 1 - cosec1 =— + 0 —, 4, 2, _ 371 -2k 7C, 4, , y=7t, , 4, , 1 +tan’*1, .2, .3, , (b) cot-12 + cot-13 = tan', , 0, , it, , 4, , ^4-, , (c) ■:, , It, , l-x2>, , _ K 1, , X, XG 0, —, , 1 + X2;, , 96., , 4, , value of /(x) is —, 4, 4171, , 7t, , it - x, x g —, rc, , .2, , Also,, , 2x, x e 0, — I, , cos----- = cos, 4 J, , 7C, , cos—, 4,, , 7t, , 4, , 2x, , 89. We know that, tan', , /w=, , Now,, , 417C^, , cos, , J, , COS -1(cosx) = X, X 6 [0,7t), , 7C ], , f, , (d) As----- = 10k + —, 4j, 4, , So,, , 2J, , Assin-1(sinx), , Hence, minimum value of /(x) does not exist and maximum, , ...., , 712, , So, area = — x it = —, 2, 2, , 1+X, , -n, — < tan, 4, , X = Jl, , x=^, 2, , As,, , Vl-x2,, , L 2J, It, Jt, xg —, n, .2 J, , f(x) G I in xg(0,3) => x =-, 1, 2, 2, , So, number of values of x are three., it + 2 tan-1 x, , ,, , x <-1, , 2tan-1 x, , ,, , -1<X<1, , - (it - 2 tan-1 x) ,, /(X), , So,, , -7t,, , x<-l, , 7t,, , x>1, , X>1, , 97. Given, /(x) = ^sin-1 x + 2 + -Jl -sin"’x, , Clearly, for domain of /(x), 1 - sin-1 x > 0, (As,sin-1x + 2 > 0,V, , xg, , [-1,1]), , sin-1x <1=>X <sinl, , Df = [—l.sinl], , Ry={-7C, 7C|, , 98. Given, /(x) = ■jsiiT1 x + 2 + -Jl-sin-1 x, where xg[- l.sinl], , 90. P = — and Q = —, 15, 15, , Lety = Vsin-1 x + 2 + 71 -sin, , So/. (Q.Nos. 91-93), A={sin-1 x}; B = {[x], {x}}; C = {[x], sin-' x}, , /(x) = A n C = sin-1 x, , Then,y > 0,Vx g[-1, sinl], y2 = (sin-1 x + 2) + (1 - sin-1 x) + 2, Now,, , •J(sin-1 x + 2)(1 - sin' X), , g(x) = B nC = [x], /i(x) = {x}, I x| is a function which is neither odd, nor discontinuous, nor, non-decreasing., , l(x) is | x|, , y2 =3 + 2, , Clearly,, , 91. l(x) = | x| is an even function, =>, , ^2. g(/(x)) = [sin-1 x], v Range of sin-1 x is - —, —, .2 2., , Range of g(/(x)) ={-2,-1,0,1}, , Also,, , 9 ( . ., — sin, 4 k, , 1, , 7, , x+ -, , 2,, , /9, -1, sin' x = — = 3 + 2,1—— 0 =3 + 3=6, Y4, 2 ., , • n /z, , x = - sin- = V6, 2j, yL(sin’1x = i)=3 + 2, , 9, 4, , ytnm(x=Sinl) = ^, , - =3, 4, , www.jeebooks.in, 93. /(/i(x)) = sin-I{x|, , =>, , Hence, range off - [V3,%/6], , K i, , Domain is R and range is 0, — I.

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www.jeebooks.in, £, , 382, , Textbook of Trigonometry, , COS ’(cos5t2) = COS-1^COS^~= y, , 4x, kx2 + 4, , 99. sin-1, , + 2 tan-1, , \, , f, , = sin-1, , cos-1(min{x, y, z})= cos-1(- 1) = 7t, , 2-^, 2, , te'-l, , 105. f(x) - tan', , = 2 tan-1 — - 2 tan-1 — = 0, 2, 2, , --1, 2___, (x, 1 + x| - + 1, 2, x, I, , 2, , r(x)= (1___+1 X2), , | x| £2 => -2 <x^2, 6x, 71, 100. cos-’ ---------- =, + 2 tan 3x, 1 + 9x2----- 2, 6x, , . _], , it, , =>, , sin-1, , =>, , . -i 2-3x, sm ---------- , = Tt - 2 tan 13x, l + (3x)2, , 6x, , 2A, , -2 +1)), 1, , 1, , ----- sin, 2, , 1 — 9x2, , 1, x2, , 1, , X, , =>, , = —— + 2tan-I3x, 2, , 2(14, I 7, , Li, , 1, =—, 2, , 2, , 9_, , 13, , 26, , 26/(1) =9, , = Tt -2tan-13x, , 1 + 9x2, 106. We have,/(x) = (tan-1 x)3 + (cot-1 x)3 = (tan-1 x + cot-1 x), , ((tan-1 x)2 - (tan-1 x)(cot-1 x) + (cot-1 x)2), , Above is true when 3x > 1, 1, => x > 3, =$x e, , 1, , 2A, , = — (tan-1 x)2 - (tan-1 x)| — - tan, 2, 2, , A, , Using cot, , /"J, , •, , 1, , sin - tan, 2, , It, , 2, , 2x, , -1, , 3n, , f, it, , tan, , x----4., , i, , 2, , _2A, rt, , 48j, Tt, , Clearly, /(x) will be minimum when tan' x----4., , . —i, , Tt It, , and, , 2*2., , x + cos, , —i, , = /(Xjmax, , z e 0, —, 2, , y + sec, , _i, , Tn, , Tt, —. It, , n n, , it, , 2, , 4, , 2, , 7C3, , 5tc, , it, , 71, , „, , [jt Y, , 5n, 5n, , lt21, 48 J, , 7?C3, , 4“ —, , X- 4, , / 2, , n +n+4, , 2, , x, , 8, , Tn = 2 tan', 1+, , The given inequation exists if equality holds, i.e., , X, n, , = 2 tan', , 2____, , x, , k n 2 + n + 4?, , x, , 1, , 2, , 2, , 3, , 107. We have, Tn = 2 arc cot, , 5?t, , TRl + T*T, , - 37t, , 32, , z S - + 7t + 7t = —, 2, 2, 7C, , /, , - = -^ = 4, a, 7?t3, , Hence,, , 2, , + 3zt = t2 - 2.J—t +-------- + 3lt, V2, 2, 2, , (, , 3rc, 2, , [tc”, , Also, t2 -, , Tt, , .. ., 3n, a=/(x)min=— I0+^- =, 48j 32, 2 X., , cos-1y 6 [0, it], , => sm, , =0, , it, , -1, , SO/. (Q. Nos. 102 to 104), , sec, , 2, , x----4, , _2\, , -i, , -i, , x = — - tan x, , 1-x2, , = sin, , 7, , +—, , and /(x) will be maximum when tan, , sin-1 x e, , It, , i, , - tan-1x, , = sin -(-rc + 2 tan 1 x) - tan 1 x, 2, , -1,, , t, , ----- tan lx, 2, , X, , 21, , 101. (x - 1) (x2 + 1) > 0 => x > 1, :., , x, , . = tan-1(x) - tan, , *<1, 2, , Here,, , x, , f, , -2tan-1 —, 2, , (, , 2_____, n(n + 1), , 2.2, , -ifn + lA, , (—J-tan, , 7, , n I, , www.jeebooks.in, LHS = RHS = —, 2, , Hence, Sn= E Tn = 2 E, n=0, , Gt", , x = 1, y = -1, z = -1 and t = J-, , n = o(, , tan'

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , 1A, 2j, , tan-1(l) - tan', , =>, , n4-1, - tan', 2, , tan', , Sn = 2 tan, , a + b = 16 =>, , Aliter: sin(30° + tan x) = —, , 2j, rn 4- I'l, , 1, , -i x ^3 •, -i x 13, => -cos(tan x) + — sin(tan x) = —, , < 2 J, , n-> ~>, , 7(1 + V^x) = 13^1 + x2, , =>, , llx2-4973x4-60 = 0, , =>, , ’ (a, , 5, , (2r 4-l)(2r - 1), J 4- (2r 4-1) (2r - 1) J, , = E tan', r=l, , 5, , = E(tan :(2r + 1) - tan-1(2r -1)) ., r=l, , = tan 11 - tan- 1 = tan, , 11-1, 1 + 11 XI, , (On squaring), , i9y/3±30y/3, x =--------22, 5^3, ---'3, 11, 5^3, x =---11, , 108. As, Ecot-1(2r2) = E tan', r=l, , 13, , i, , n., , Km Sn = 2. — = n = kK (given) => k = 1, , r=l, , a^3, ----b, a+b, 2 ) =8, , 5^3, x =---11, , Sn = tan' j -1 - tan-1(0), , 383, , (Reject), , 111. Clearly x2‘ -4x + a >or>-X, V, V3, , xgR, , y~*, ( 5, . , 1, (, tan E cot (2r ) = tan tan, V =1, , Now, tan, , /, , ( 5, \ 5x 4- 6, _5, E cot-1(2r2) = ------Ir=i, J 6x4-5----- 6, , 5, 6, , (of), ■+, , 5x 4-=6 ------6x4-5, , y=0, , Graph ofy=cot"1x, , => The given equations has no real solution., 109. min(x2 4- 4x4-7) =3, , 0 (0. 0), , x2-4x + a + -L>0,VxgR, V3, So, discriminant <0 => 16 - 4^a +, <0, , =>, , It2, max- -—,3 = 3, 4, , lim 3., o, , sin z, =3, z, , 1, 1, 4-a—z= <0 => a>4—t=, •V3, , 7t, -1 |, 110. sin —, 4- tan x =, .6, ), %, , a e I 4 - 1-t=, °°1, < V3 J, , _i, , Hence, minimum integral a = 4, , — + tan x = sm, 6, , 112. L :34-[3Q(24 - 2) 4-3C2] = 36, , MfIfx>0,Sgn(x) = l, , tan- x = tan, , f(x) = 0 + — = —, , 14,, , 13, , 3^3, , = tan', , 13___ 1_, 3^3 J3, 13, 1+—, 9, , . J 10 ( 9 ) 1, , ,an, , 2, For x = 0 f(x) is not defined, , 2, , :. for x < 0, /(x) = n — = —, 2 2, M=1, N: Coefficient of ts = coefficient of t2 ir<l + t2)5 x coefficient, , oft3in(l + t3)8, , www.jeebooks.in, tan- x = tan', , 5^3, , 11, , Hence, =>, , = 5X8 = 40, L = 36; M = 1 and N = 40, N-LM = 40-36 = 4

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www.jeebooks.in, ■, , 384, , Textbook of Trigonometry, , 113., , Ma, , = tan-1, , 2, , sec 2(u) + sec4(v) + sec6(w) e [3,«»), , n(sec2u + sec4 v + sec6w) e [3n,°°), cos-1 x + cos-Iy + cos-‘z e [0, 3tt], , But, , So equation is possible of LHS = RHS = 3tt, cos-1 x = cos"1 y = cos-1 z = K, , = sin- ‘(sin(4n - (4n - 12))), , + cos-‘(cos(4lC —(4k -12))), =-(47t - 12) + (47t - 12) = 0, So that, (n - 2)x2 + 8x + n + 4 > 0, V x 6 R, , =>, and, , n -2 > 0 => n >3, 82 - 4(n - 2) (n + 4) < 0, , or,, , n2 + 2n - 24 > 0 => n = 4, n > 5 => n = 5, , 118. 1 + sin(cos- ’ x) + sin2(cos-1 x) +...«» = 2, , x = y=z =-l, sec2 u = sec4 v = sec6 w = 1, , and, , =>, , --------------- ~ = 2, 1 - sin (cos x), , u = 7t, v = 2lt, w = lit, .2000, , + y2002, , 2004, , 3671, , +------------ = 1 + 1 + 1 + — = 9, U+V+w, 67t, , 115. Given, /(x) = cos(tan-‘(sin(cot-‘ x))), /, /, = cos tan-1 sin-1, \, /, -1, , tan 1 -r, , = COS, , \, /, - cos, , COS, , _1, , 1__, , \\, , - = 1 -sin(cos-1x), , 2, , a/3, , or, , x = — orl2x =9, 2, , 119. Here | cos x | = sin-1 (sin x), , i, , y, , IV1 + x, , yj, , .2, ( V111 + x', A, , H, , -1—, , or, , or sin(cos-1 x) = - or cos- ‘ x = —, 2, 6, , •Ji + x.2, , i1, , I, , 117. We have, sin-‘(sin 12) + cos-‘(cos 12), , Number of solutions are three., 114. sec2 u, sec4 v, sec6 w e [1,00), , i, , J, , Hence, the value of X is equal to 2., , 15, , 0, , cos2asec2p + cos2Psec2a) 1, , —x, , ■■, , '2+ x2)), , _ 71+_ [ x2 + 1, , y=cosx, n/2~, , -n/2, , 1/2, , ^2 + x2 I x2 + 2, y, , A = 1, B = 2, A + B = l + 2=3, , 116. We have,, RHS = tan-1 {tan2(a + P) • tan2(a - P)} + tan-11, = tan !-, , - tanL, = tan”, , 1 - tan2(a + p) tan2(a - p), , sin2(a + P)sin2(a - P) + cos2(a + P) cos2(a - p), , --------------- —--------------------------------------------------------------------------------------------------- ►, , cos2(a + p)cos2(a - P) - sin2(a + P) sin2(a - p), {2sin(a + P)sin(a - P)}2, , {2cos(a + P)cos(a ~P)}2, , - {2sin(a + P)sin(a - P)}2, (cos2p - cos2a)2 + (cos2a 4- cos2P)2, , (cos2a + cos2P)2 -(cos2P - cos2a)2, , n, , x, , -n, , tan2(a + P)tan2(a ~P) +1, , + {2 cos(a + p) cos(a - p)}2, , = tan-1, , -n/2, , J, , ., , 7t, , If---- < x < — then cosx = x, 2, 2, In the case there is one solution obtained graphically,, it, , If — < x < it then - cosx, 2, = sin-1 {sin(n - x)} = it - x, , cosx = x-7t, In this case there is one solution, obtained graphically., 7C, , If- it <, x <---- then, 2, - cosx = sin-1 {sin(- it - x)} = - x - it, , www.jeebooks.in, = tan-1-, , cos2a + cos22p, 2cos2acos2p, , i.e.,, cosx = x + n., This gives no solution as can be seen from then graphs.

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www.jeebooks.in, s, , Chap 04 Inverse Trigonometric Functions, , 385, , i, , 120. From graph, , =, , =>, , \ •, , Statement I is true, Statement II is true, Statement II is a, correct explanation for Statement I., , ft, -+, , +n, , 3rc, 2., , x, , 124. sin-12x + sin-,3x = —, 3, , 0, , n, , 5, , 2, , Tt, , Graph for tan"1 (tan x), , 0, , 2rt, , k, , 19x2 = -=>x = ± E, 4, , >X, , /T., I, 1 K, x - J— is only solution., , Statement II is true and correct explanation for Statement 1., X, , -i, , 7A7, , _i, , > tan x > tan y, , V'X>y,, , V76, , But sum of two negative angles cannot be j-, , Graph for cos-1 (cos x), , 121. sin 1x = tan-1 ■, , Tt, , 2, , A/W\, -Tt, , —t, , Tt, , 6x2 + - • = 1 - 13x2 + 36x4, 2, , I, —2lt, , _], , .’.---- cos 2x +----- cos 3x = —, 2, 2, 3, 2n, cos -u2x+cos-i,3x = —, 3, 1, 6x2 - - 13x2 + 36x* = 2, , 3rt, 2, , 125. Statement 2 is correct, from which we can say, cot-1 x + cos"12x = - Tt is not possible. Hence, both the, statements are correct, and Statement 2 is the correct, explanation of Statement 1., , 126. (A) In (0, cos 1), we have cos-1 x > sin-1 x, Statement II is true, , Also, cos-1 x > 1 and sin-1 x < 1, , e < 7t, 1, Ve Vk, , The greatest is (cos-1 x).COS, 1, , 1, , = tt and the least is, , (sin-1 x)co$' X=h, , by Statement II, , and, , sin 1, , X, , > tan-1, , (sin-,xr‘*x-<(cos-1x) sin"1 r, , h<h, , > tan 1, , So,, Statement I is true, , 1, (B) Similarly, in cost < x < —j=, , 122. cosec-'x >sec-1 x, , V2, , -i, , *, , -i, , *, , -i, , cosec x >---- cosec x, 4, , cosec x> —, 4, , and, , T, 1 ', - + ~7=, , 2 Ji., , g, , [i, -Ji], , Statement II is true and explains Statement I., , 123. /(x)=sin-1 f 2x, , V + x2, , = n - 2 tan-1 x, x > 1, , rw=- 1 +2 x2, , cos-1 x > sin-1 x and both are less than 1 So, greatest is t3 and, least is t2 and t4 >, Hence,, t3 >, > t2, , (C) For-4= < x <sinl, V2, We have, 1 > sin-1 x > cos-1 x, , So, greatest is t2 and least is t3 also r, > t4, Hence,, t2> >t4 >t3, (D) For sinl < x < 1, we have, sin-1x > 1 > cos-1x, , So, the greatest is and least is t3 and t2 > t4, Hence,, > t2 > t4 > t3, , www.jeebooks.in

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www.jeebooks.in, 386, , Textbook of Trigonometry, , 127. Follow the following graph for solution, X, , <=>, , x > 12 and cos-1, , "7x2 -25, x, , y=sin-1x J, , -1--, , Jx2 -25, , 12, , X, , X, , <=>, , x > 12 and, , <=>, , x>12and x2-25 = 122, , I, , (A), , 4^2, , x > 12 and x2 = 169, , y=-x, , ;, , x>12andx = ± 13, , -1--, , 13 is only solution., , I, i, I, , 129. Taking tan on both sides, x+1x-1, , ■4;, , I, , 1, , x -1, X, = - 7 on simplification, we get x = 2, x +1 x-1, , X, , x-1, , y- cos-1x, Substituting, we get, , <2--, , I, I, I, , tan-13 + tan"1, , y=x, , (B), , = tan- ’(- 7) but L.H.S is, , it + tan-’(-7). Hence, ho solution., 1 --, , 130. Let X =, XY =, , a(a + b + c), -,y =, be, , b(a + b + c), , ac, , c(a + b + c), , ,Z =, , ab, , a(a + b + c) b(a + b + c)_a + b + c, be, , c, , ac, , ., a+b, = 11 +------- >l(a,b,c> 0), c, , tan"1X + tan-17 = k + tan-, , ! X+ Y, 1-XY, , la(a + b + c), , be, !, , = it + tan"1 N, , b(a + b + c), , V, , ac, , a+ b+c, c, , . x, , la + b + c, , a, + — + -7=, N ___________, c, a, b_, (a + b), , = n + tan 1, x, /, , ■, , = % + tan"1 -, , c, , 2—, <, , 14-, , ab, , 1, , 1, , 1, , 1, , A, , (C), , c, • 7, ------ ;------ /, .\ /, a + b + c( a + b], , y='cor1x !, -------, , -it - tan 1, , yjc(a + b + c), , x, , /, , = it - tan-1 Z, , ab, , tan"1X + tan-1 y + tan-1 Z = it, , 1, , 2, , *X, , 131. Let 0 = cosec"’^(n2 + 1) (n2 + 2n + 2), cosec2© = (n2 + 1) (n2 + 2n + 2), , = (n2 + I)2 + 2n(n2 + 1) + n2 + 1 = (n2 + n + 1) + 1, 128. sin-, , cot20 =(n2 + n + I)2, , x‘ -25, , -■■("I- 2, , It, , tanO =, , (n + 1) - n, , 1, , www.jeebooks.in, <=>, , x > 12 and cos-1, , X, , x, , n2 + n + 1, , 0 = tan-1, , 1 + (n + l)n, , (n + 1) - n, , 1 + (n + l)n

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www.jeebooks.in, 388, , Textbook of Trigonometry, , 28 - e,28, = isin9, i(e-2B + e2B), ■2B, , e2B, e2B, , or, , -28, , _, , 139. We have,, , Let, , 0<x<l, cot-' x = 9 => cot 9 = x, , n, , 0>, , 4, , 2J, , and, , i<7t, , cos 0 =, , O'!, , 1, , (p2-p + p;>o, , 4s”2 afp2 —p +2), , 2, , <1, , B, , -.(i), , A, , x, , Now, -^1 + x.2‘ [{x cos (cot-1 x) + sin (cot-1 x)}2 -1]il/2, 1, /, , 1, 1, +->2), 4 4, , k, , = cos (cot-1 x), , - P + - = 0 have, 2, , i4.p2-P+l = (p-l'l, , X, , 7i + x.2, , 1, , real roots., , But 4s”2 a, , = sin (cot-1 x), , .2, , C, , X2 + 2x+ ^P2, , 137. The quadratic equation 4 sec2 a, , 1, , sin 9 =, , B = - -In tan------2, .4 2J, , =>, , >n, , cos(a + p) > 0, , 2., , => discriminant = 4 - 4.4s” “, , 3, , 3rt, , 9, , 1., , 2, , 2, , sin2 f K, U, , =>, , n, , Now, a + P is slightly greater than —., , <4 _2, 9, , = cot, , 1, , cosP < 0 and sinP < 0, , 1 - cos I — - 9, U, , cos2I! n, , e2B, , >cos, , 9, , 9, , 1 + cos I — - 9, <2, , 1 + sin0, 1 - sin9, , 4, , 4, , P =3 cos, , = sin9, , ■2B, , :., , or, , 4 1, -<- => cos, 9 2, , As, , 2, , 1, , =7177, , < 7i + *: Vi+<, .2, , .2, , i.e. equation will be satisfied only when, , 1/2, , \z, -1, , -il/2, , 4^“ = 4 andp2-p + l = l, , 1 + x2, , = -Jl + x2, , 4, , 2, , -1, , 2, , sec2 a = 1 and I B - - I, V, 2), , =>, , =0, , = 71 + X2[l + x2-l],/2 = xVl, , cos2a = 1 and B = 2, , 140. Given, tan-1y = tan-1 x + tan, , a = nn and p = -1, 2, , =>, , cosa + cos-Ip = cosnn + cos', , =>, , 1 + y,, , when n is even integer, , -1 + ~,, , when n is odd integer, , tan- y = tan, , 1 -x, , cosa + cos pis---- 1 and — + 1., , 138. Here, a =3 sin, , and P =3 cos, , sin', , -if 6>, 1117, , ., , 3, , 6 1, as— >9, 11 2, n, 6, , 2x, J-x2, , x+y, , where x > 0, y > 0 and xy < 1], f x - x3 + 2x', = tan' [1 - x2 -2x\, , , 7t, , 3, , 7s, , [ v tan 1 x + tan-1y = tan, , i.e. the value of;, —i, , 1, 2x, , where |x| <, J-x2,, , 2x, 1 -xz, , 1, , 2, , .2, , 4, , (6 > n, a = 3sin — >— => cosa < 0, 111? 2, , tan- y = tan', =>, , <3x-x3>, J-3x2,, , 3x-x3, , ^777, , www.jeebooks.in, Now,, , P = 3 cos, , Aliter, , . ., , 1, V3, , X<-7=, , 1, , 1, , => — —7= < X < —7=, , v3, , -J3

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www.jeebooks.in, Chap 04 Inverse Trigonometric Functions, , it n n, ----- <0 < —, 6, 6, tan 1 y = 0 + tan-1 (tan 20), , =>, , y2 = xz, , =>, , Let x = tan 0, , 143., , = 0 + 20=30, y = tan 30, 3 tan 0 - tan30, y =------------ —, 1-3 tan20, , =>, , Since, x, y and z are in an AP as well as in a GP., x =y = z, 3, 5, = tan, Since, cosec, 4, 3, 5, , 3x-x3, , k=l, , ., , n=l, , cot tan, , \', , 2, 3., , -1 3, , — + tan, 4, , 1+ £2fc, , 141. We have, cot y cot, , 3 2, -+4 __ 3 = cot tan, 1, 1 --, , J, = cot tan, , 23, , => cot y cot ’(1 + 2 + 4 + 6 + 8+ ... + 2n), 23, , ©H, , = cot tan, , => cot y cot-1 {1 + n(n + 1)}, n=1, , cot, , 144. We have,, , Xtan', , n=1, , n =1, 23, , cot y (tan-1(n + 1) - tan-1 n), , =>, , x, + sin, 5., , sin, , (n + 1) - n, 1 + n(n + 1), , cot y tan, , =>, , sin', , =>, , sin, , =>, , sin', , x, 5., , It, . •, ---- sin, 2, [4, x, = cos, <5, 5., , n =1, , => cot [(tan-12 - tan-1l)+ (tan-13 — tan-12), , = cot cot, , 25, 23, , cos, 25, 23, =>, , 2y = x + z, Also, tan-1 x, tan"1 y and tan-1 z are in an AP., , 2 tan-1y = tan-Ix+ tan-1(z), =i>, , =>, , if 2y ', , tan-1, , x+z, 1-/, , = tan-1, , x+z, 1 - xz, , x+z, 1 -xz, , it, , 2, 5/, , x=3, , cos, , 145. Given that,, , 142. Since, x, y and z are in an AP., , n, 2, , -15), , sin, , + (tan-14 - tan-13)] + ... + (tan-124 - tan-123)], , => cot(tan-124 - tan-11), /, 24-1, f, -123, =» cot tan, = cot tan —, 1 + 24(1), I, 25, , 5), .4, 4, 5, , x, + cosec, 5,, , sin, , 1, 1 + n(n + 1), , 23, , ©, , 2J, , n=1, , 23, , 3, , 4, , f, , 23, , 389, , =>, , ^ + 7T7, 2, , .2, , 2, , - =a, , x - cos, , 2, , -Z1, 4, , =a, , /, , I - — = cos a, 4, , rv, , 2^1 - x'.2 II, , =2 cos a -xy, , 4, , On squaring both sides, we get, ———— = 4cos2 a + x2y2 - 4xv cos a, 4, cos a, 4 - 4x2 -y2 + x2y2 = 4cos2 a + x‘, 2*2, , 4x - 4xy cos a + y - 4sin a, , www.jeebooks.in

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www.jeebooks.in, Skills in Mathematics for, , EE MAIN & ADVANCED, , Trigonomet, With Sessionwise Theory & Exercises, It is generally seen that students leave the trigonometry class believing that they have, mastered the concepts, but when it comes to actual problem solving, they mess up all the, things. The book in your hand just aims to bridge this gap. The book completely covers the, trigonometry of J EE Main & Advanced, guiding the students step by step from the right, triangle to the definitions of trigonometric functions and their properties. The treatment of, the subject matter is methodological and designed to involve the students completely., This hands-on text of trigonometry with its updated contents, vividly covers the topics of, basic trigonometry viz., trigonometric functions and identities, trigonometric equations and, inequations, properties and solutions of triangles and the inverse trigonometric functions., Prescribedfor those aiming to appear in JEE & other Entrances and also to the ones who wish to have, the complete appreciation of the trigonometricfunctions.., , ABOUTTHEAUTHOR, Amit M. Agarwal is a renowned name and an authority in the field of Mathematics with a, number of accolades to his credit. Immediately after doing his Post Graduation from Meerut, University in 1997, Mr. Agarwal started teaching Mathematics in his own institute at Meerut, and simultaneously writing the books for JEE. Amit sir's students has always been among the, good JEE Ranks. For the last 20 years he is continuously teaching JEE aspirants and writing &, revising the books. Presently he is teaching in Mumbai., , BOOKS IN THE SERIES, , ^arihant, Arihant Prakashan (Series). Meerut I1, , IIIHIIHIIIIIIII 13, ARlHANT, , 9H/893 1 2Hl 46927, , www.jeebooks.in, Code : B017, , ?255.00