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Chapter, , 2, , INVERSE TRIGONOMETRIC, FUNCTIONS, 2.1 Overview, 2.1.1 Inverse function, Inverse of a function ‘f ’ exists, if the function is one-one and onto, i.e, bijective., Since trigonometric functions are many-one over their domains, we restrict their, domains and co-domains in order to make them one-one and onto and then find, their inverse. The domains and ranges (principal value branches) of inverse, trigonometric functions are given below:, Functions, Domain, Range (Principal value, branches), , – , ,, 2 2, [0,π], , y = sin–1x, , [–1,1], , y = cos–1x, , [–1,1], , y = cosec–1x, , R– (–1,1), , y = sec–1x, , R– (–1,1), , [0,π] –, , y = tan–1x, , R, , – , ,, 2 2, (0,π), , y = cot–1x, , – , , – {0}, 2 2, , , 2, , R, Notes:, (i) The symbol sin–1x should not be confused with (sinx)–1. Infact sin–1x is an, angle, the value of whose sine is x, similarly for other trigonometric functions., (ii) The smallest numerical value, either positive or negative, of θ is called the, principal value of the function., , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , 19, , (iii) Whenever no branch of an inverse trigonometric function is mentioned, we mean, the principal value branch. The value of the inverse trigonometic function which, lies in the range of principal branch is its principal value., 2.1.2 Graph of an inverse trigonometric function, The graph of an inverse trigonometric function can be obtained from the graph of, original function by interchanging x-axis and y-axis, i.e, if (a, b) is a point on the graph, of trigonometric function, then (b, a) becomes the corresponding point on the graph of, its inverse trigonometric function., It can be shown that the graph of an inverse function can be obtained from the, corresponding graph of original function as a mirror image (i.e., reflection) along the, line y = x., 2.1.3 Properties of inverse trigonometric functions, 1., , –π π, ,, 2 2, , sin–1 (sin x) = x, , :, , x∈, , cos–1(cos x) = x, , :, , x ∈[0, π ], , tan–1(tan x) = x, , :, ,  –  , x ∈ , ,  2 2, , cot–1(cot x) = x, , :, , x ∈ (0,  ), , sec–1(sec x) = x, , :, , x ∈[0, ] –, , cosec–1(cosec x) = x :, , , 2, , – , , – {0}, 2 2, x ∈[–1,1], x ∈[–1,1], x ∈R, x ∈R, x ∈R – (–1,1), x ∈R – (–1,1), x∈, , 2., , sin (sin–1 x) = x, cos (cos–1 x) = x, tan (tan–1 x) = x, cot (cot–1 x) = x, sec (sec–1 x) = x, cosec (cosec–1 x) = x, , 3., , sin –1, , 1, = cosec –1 x :, x, , x ∈R – (–1,1), , cos –1, , 1, = sec –1 x, x, , x ∈R – (–1,1), , :, :, :, :, :, :, , :, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , –1, , Solution If cos, , 3, 2 = θ , then cos θ =, , 3, ., 2, , Since we are considering principal branch, θ ∈ [0, π]. Also, since, , the first quadrant, hence cos–1, , 3, 2, , Example 2 Evaluate tan–1 sin, , –, 2, , Solution tan–1 sin, , –, 2, , ., , ,   , , = tan–1  − sin    = tan–1(–1) = − ., 2,  , 4, , , 13, ., 6, , 13, π , , , –1 , = cos–1  cos (2π + )  = cos  cos , 6, 6 , 6, , , =, , π, ., 6, , Example 4 Find the value of tan–1 tan, Solution tan–1 tan, , 3, > 0, θ being in, 2, , , ., 6, , =, , Example 3 Find the value of cos–1 cos, Solution cos–1 cos, , 21, , 9, 8, , 9, ., 8, , π, , = tan–1 tan  π + , 8, , ,  π , , –1 , = tan  tan    =,  8 , 8, , –1, Example 5 Evaluate tan (tan (– 4))., , Solution Since tan (tan–1x) = x, ∀ x ∈ R, tan (tan–1(– 4) = – 4., Example 6 Evaluate: tan–1 3 – sec–1 (–2) ., , 20/04/2018

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22, , MATHEMATICS, , Solution, , tan–1 3 – sec–1 (– 2) = tan–1 3 – [π – sec–12], =, , π, 2π π, π, 1, − π+ cos –1   =− + = − ., 3, 3 3, 3,  2, , 3, –1, –1, Example 7 Evaluate: sin cos sin 2, –1, –1, Solution sin cos sin, , 3, 2, , = sin –1 cos, , ., , , 3, , –1, = sin, , 1, , = ., 2, 6, , Example 8 Prove that tan(cot–1x) = cot (tan–1x). State with reason whether the, equality is valid for all values of x., Solution Let cot–1x = θ. Then cot θ = x, or, tan, , , , –  = x ⇒ tan –1 x = – , 2, 2, , , , , –1, –1 , –1, So tan(cot x ) = tan  = cot  –   = cot  − cot x  = cot(tan x ), 2, 2, , , , , The equality is valid for all values of x since tan–1x and cot–1x are true for x ∈ R.,  –1 y , Example 9 Find the value of sec  tan, ., 2, , –1, Solution Let tan, , which gives, , Therefore,, , y, y,   , = , where  ∈  − ,  . So, tanθ = ,, 2, 2,  2 2, , sec=, , 4 + y2, ., 2, , 4 + y2, y, , sec  tan –1  = sec =, ., 2, 2, , , –1, Example 10 Find value of tan (cos–1x) and hence evaluate tan cos, , 8, ., 17, , Solution Let cos–1x = θ, then cos θ = x, where θ ∈ [0,π], , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , tan(cos–1x) = tan  =, , Therefore,, , 23, , 1 – cos 2 , 1 – x2, =, ., cos, x, 2, ,  8, 1–  , 8, 15 ., ,  17 , =, tan  cos –1  =, 8, 17 , 8, , 17, , Hence, , –1, Example 11 Find the value of sin 2cot, , –5, 12, ,  –5 , −5, Solution Let cot–1   = y . Then cot y =, ., 12,  12 , –1, Now sin 2cot, , –5, 12, , = 2, , = 2siny cosy, , =, , = sin 2y, , 12, 13, , ,   ,  since cot y < 0, so y ∈ 2 ,   , , , , , –5, 13, , –120, 169, , –1, Example 12 Evaluate cos sin, , –1, Solution cos sin, , –1, = cos sin, , 3, 1, = 4 1– 4, , 1, 4, + sec –1, 4, 3, , 1, 4, 3,  –1 1, + sec –1, + cos –1 , = cos sin, 4, 3, 4, 4, , , 1, 3, 1, 3, cos cos –1, – sin sin –1 sin cos –1, 4, 4, 4, 4, 2, , –, , 1, 3, 1–, 4, 4, , 2, , 3 15 1 7 3 15 – 7, = 4 4 –4 4 =, 16, , ., , 20/04/2018

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24, , MATHEMATICS, , Long Answer (L.A.), Example 13 Prove that 2sin–1, Solution Let sin–1, Thus tan θ =, Therefore,, , 3, 17, π, – tan–1, =, 5, 31, 4, ,  −π π , 3, 3, = θ, then sinθ = , where θ ∈  , , 5, 5,  2 2, , 3, 3, , which gives θ = tan–1 ., 4, 4, 2sin–1, , 3, 17, – tan–1, 5, 31, , = 2θ – tan–1, , 17, 31, , = 2 tan–1, , 3, 17, – tan–1, 4, 31, ,  3 ,  2. , 17, 24, 17, tan –1  4  – tan –1, − tan –1, =, 31 = tan–1,  1– 9 , 7, 31,  16 ,  24 17 , −, , , π, tan –1  7 31 , =, =,  1 + 24 . 17 , 4, 7 31 , , Example 14 Prove that, cot–17 + cot–18 + cot–118 = cot–13, Solution We have, cot–17 + cot–18 + cot–118, = tan–1, , 1, 1, 1, + tan–1 + tan–1, 7, 8, 18, ,  1 1 ,  + , 1, tan  7 8  + tan –1, =, 1, 1, 18,  1− × ,  7 8, , (since cot–1 x = tan–1, , –1, , (since x . y =, , 1, , if x > 0), x, , 1 1, . < 1), 7 8, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, ,  3 1 , +, , , tan –1  11 18 , –1 3, –1 1, + tan, = tan, =,  1− 3 × 1 , 11, 18,  11 18 , –1, = tan, , 65, 195, , –1, = tan, , 1, 3, , 25, , (since xy < 1), , = cot–1 3, , Example 15 Which is greater, tan 1 or tan–1 1?, Solution From Fig. 2.1, we note that tan x is an increasing function in the interval, ,  −π π , π, π, ⇒ tan 1 > tan . This gives,  ,  , since 1 >, 4, 4,  2 2, , Y, , tan x, , tan 1 > 1, ⇒, ⇒, , π, 4, tan 1 > 1 > tan–1 (1)., , tan 1 > 1 >, , O, , –/2, , X, /4, , /2, , Example 16 Find the value of, , 2, , sin  2 tan –1  + cos(tan –1 3) ., 3, , Solution Let tan–1, , Therefore,, , 2, = x and tan–1, 3, , 3 = y so that tan x =, , 2, and tan y =, 3, , 3., , 2, , sin  2 tan –1  + cos(tan –1 3), 3, , = sin (2x) + cos y, 2, 2., 1, 2 tan x, 1, 3 +, +, 2, = 1 + tan x, =, 4, 1+ tan 2 y, 1+, 1+, 3, 9, , ( ), , =, , 2, , 12 1 37, + = ., 13 2 26, , 20/04/2018

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26, , MATHEMATICS, , Example 17 Solve for x, ,  1− x  1, –1, tan –1 ,  = tan x, x > 0, +, 1, x, 2, , , Solution, ,  1− x , –1, From given equation, we have 2 tan –1 ,  = tan x,  1+ x , , ⇒, , 2  tan –1 1 − tan –1 x  = tan –1 x, , ⇒, ,  π, π, 2   = 3tan –1 x ⇒, = tan –1 x, 6, 4, , ⇒, , x=, , 1, 3, , Example 18 Find the values of x which satisfy the equation, sin–1 x + sin–1 (1 – x) = cos–1 x., Solution From the given equation, we have, sin (sin–1 x + sin–1 (1 – x)) = sin (cos–1x), ⇒ sin (sin–1 x) cos (sin–1 (1 – x)) + cos (sin–1 x) sin (sin–1 (1 – x) ) = sin (cos–1 x), , ⇒ x 1– (1– x) 2 + (1− x ) 1 − x 2 = 1− x 2, ⇒ x 2 x – x 2 + 1− x 2 (1− x −1) = 0, ⇒x, , (, , ), , 2 x – x2 − 1− x2 = 0, , ⇒x = 0, , or, , 2x – x2 = 1 – x2, , ⇒x = 0, , or, , x=, , 1, ., 2, , Example 19 Solve the equation sin–16x + sin–1 6 3 x = −, , π, 2, , π, –1, Solution From the given equation, we have sin–1 6x = − − sin 6 3 x, 2, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , ⇒, ,  π, , –1, sin (sin–1 6x) = sin  − − sin 6 3 x ,  2, , , ⇒, , 6x = – cos (sin–1 6 3 x), , ⇒, , 6x = – 1−108x 2 . Squaring, we get, , 27, , 36x2 = 1 – 108x2, ⇒, , ⇒ x= ±, , 144x2 = 1, , 1, 12, , Note that x = –, , 1, 1, is the only root of the equation as x =, does not satisfy it., 12, 12, , Example 20, , Show that,  α,  π β , –1 sin α cos β, 2 tan–1  tan .tan  −   = tan, 2, cos α + sin β,  4 2 , , , α,  π β, .tan  − , 2,  4 2, –1, Solution L.H.S. = tan, α,  π β, 1− tan 2 tan 2  − , 2,  4 2, 2 tan, , , –1, –1 2 x ,  since 2 tan x = tan, , 1− x 2 , , , β, 1− tan, α, 2, 2 tan, 2 1 + tan β, –1, 2, = tan, 2, β, , 1, tan, −, α , 2, 1− tan 2, , , 2  1+ tan β , 2, , , β, , .  1− tan 2 , 2, , = tan –1, 2, 2, β, β, , 2 α , 1, tan, tan, 1, tan, +, −, −, , , , , 2, 2 , 2, , 2 tan, , α, 2, , 20/04/2018

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28, , MATHEMATICS, , =, , , 2 β,  1− tan , 2, , tan –1, β , , 2β , 2 α, 2 α,  1+ tan   1− tan,  + 2 tan  1+ tan, , 2 , 2, 2 , 2, , , =, , α, β, 1− tan 2, 2, 2, 2 α, 2β, 1 + tan, 1+ tan, 2, 2, tan –1, β, 2 α, 1− tan, 2 tan, 2+, 2, 2 α, 2β, 1 + tan, 1+ tan, 2, 2, , =, ,  sin α cos β , tan –1 , ,  cos α + sin β , , 2 tan, , α, 2, , 2 tan, , = R.H.S., , Objective type questions, Choose the correct answer from the given four options in each of the Examples 21 to 41., Example 21 Which of the following corresponds to the principal value branch of tan–1?, ,  π π, (A)  − , ,  2 2, , (B), ,  π π,  − 2 , 2 , ,  π π, (C)  − ,  – {0},  2 2, , (D) (0, π), , Solution (A) is the correct answer., Example 22 The principal value branch of sec–1 is, (A), ,  π π,  − 2 , 2  − {0}, , (C) (0, π), , (B), , π, , 2, , [0, π] − , ,  π π, (D)  − , ,  2 2, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , 29, , Solution (B) is the correct answer., Example 23 One branch of cos–1 other than the principal value branch corresponds to, ,  π 3π , (A)  , , 2 2 , (C) (0, π), , (B), , 3π , , 2, , [π , 2π]− , , (D) [2π, 3π], , Solution (D) is the correct answer.,  43π  , –1 , Example 24 The value of sin  cos ,   is,  5 , , , 3π, 5, , (A), , (B), , −7 π, 5, , (C), , π, 10, , (D) –, , π, 10, , 40π+ 3π , 3π , , –1 , –1, Solution (D) is the correct answer. sin  cos,  = sin cos  8π+ , 5, 5 , , , , 3π ,  π 3π  , –1 , –1 , = sin  cos,  = sin  sin  −  , 5 , ,   2 5 , –1 , = sin  sin, , , π,  π , −  = − ., 10,  10  , , Example 25 The principal value of the expression cos–1 [cos (– 680°)] is, (A), , 2π, 9, , (B), , − 2π, 9, , (C), , 34π, 9, , (D), , π, 9, , Solution (A) is the correct answer. cos–1 (cos (680°)) = cos–1 [cos (720° – 40°)], = cos–1 [cos (– 40°)] = cos–1 [cos (40°)] = 40° =, , 2π, ., 9, , Example 26 The value of cot (sin–1x) is, (A), , 1+ x2, x, , x, (B), , 1+ x 2, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , (A) −, , 2π, 3, , (B), , −, , π, 3, , (C), , 4π, 3, , (D), , 31, , 5π, ., 3, , Solution (B) is the correct answer., − 3, π, π, π, –1 , –1 , sin –1 ,  = sin  – sin  = – sin  sin  = – ., 3, 3, 3, , ,  2 , , Example 30 The greatest and least values of (sin–1x)2 + (cos–1x)2 are respectively, (A), , 5π 2, π2, and, 4, 8, , (C), , π2, −π2, and, 4, 4, , (B), , π, −π, and, 2, 2, , (D), , π2, and 0 ., 4, , Solution (A) is the correct answer. We have, (sin–1x)2 + (cos–1x)2 = (sin–1x + cos–1x)2 – 2 sin–1x cos–1 x, =, , π2, π, − 2sin –1 x  − sin –1, 4, 2, , =, , π2, − π sin –1 x + 2 sin –1 x, 4, , , x, , , (, , ), , 2, , 2 π, , π2 , –1, –1, 2, sin, x, −, sin, x, +, , = , 2, 8 , , , (, , ), , 2,  –1, π  π2, 2, sin, x, −,  +, = , 4  16, , , , ., , ,  −π π  2 π2 ,  π2  π 2, 2, 2, i.e., and the Greatest value is  2 − 4  + 16  ,, Thus, the least value is  , 8, , ,  16 , , , i.e., , 5π2, ., 4, , Example 31 Let θ = sin–1 (sin (– 600°), then value of θ is, , 20/04/2018

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32, , MATHEMATICS, , π, 3, , (A), , (B), , π, 2, , (C), , 2π, 3, , (D), , − 2π, ., 3, , Solution (A) is the correct answer., , π , ,  −10π , –1, sin –1 sin  − 600 ×,  = sin sin , , 180 , ,  3 , 2π  , 2π , , –1 , –1 , = sin  − sin  4 π −   = sin  sin , 3 , 3 , , , , , π , π π, , –1 , –1 , = sin  sin  π −   = sin  sin  = ., 3 , 3 3, ,  , , Example 32 The domain of the function y = sin–1 (– x2) is, (A) [0, 1], (C) [–1, 1], , (B) (0, 1), (D) φ, , Solution (C) is the correct answer. y = sin–1 (– x2) ⇒ siny = – x2, i.e. – 1 ≤ – x2 ≤ 1 (since – 1 ≤ sin y ≤ 1), ⇒ 1 ≥ x2 ≥ – 1, ⇒ 0 ≤ x2 ≤ 1, ⇒, , x ≤ 1 i.e. − 1 ≤ x ≤ 1, , Example 33 The domain of y = cos–1 (x2 – 4) is, (B) [0, π], , (A) [3, 5], (C), ,  − 5, − 3  ∩  − 5, 3 , ,  , , , (D), ,  − 5, − 3  ∪  3, 5 , ,  , , , Solution (D) is the correct answer. y = cos–1 (x2 – 4 ) ⇒ cosy = x2 – 4, i.e. – 1 ≤ x2 – 4 ≤ 1 (since – 1 ≤ cos y ≤ 1), ⇒ 3 ≤ x2 ≤ 5, ⇒, ⇒, , 3≤ x ≤ 5, , x∈ − 5, − 3  ∪  3, 5 , , Example 34 The domain of the function defined by f (x) = sin–1x + cosx is, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , (B) [–1, π + 1], , (A) [–1, 1], , ( – ∞, ∞ ), , (C), , 33, , (D) φ, , Solution (A) is the correct answer. The domain of cos is R and the domain of sin–1 is, [–1, 1]. Therefore, the domain of cosx + sin–1x is R ∩ [ –1,1] , i.e., [–1, 1]., Example 35 The value of sin (2 sin–1 (.6)) is, (A) .48, , (B) .96, , (C) 1.2, , (D) sin 1.2, , Solution (B) is the correct answer. Let sin–1 (.6) = θ, i.e., sin θ = .6., Now sin (2θ) = 2 sinθ cosθ = 2 (.6) (.8) = .96., Example 36 If sin–1 x + sin–1 y =, (A), , π, 2, , π, , then value of cos–1 x + cos–1 y is, 2, , (B) π, , (C) 0, , Solution (A) is the correct answer. Given that sin–1 x + sin–1 y =, Therefore,, ⇒, , (D), , 2π, 3, , (D), , 3, 4, , π, ., 2, , π, π, –1 , –1  π,  – cos x  +  – cos y  =, 2,  2,  2, , cos–1x + cos–1y =, , π, ., 2, , 1, , –1 3, + tan –1  is, Example 37 The value of tan  cos, 5, 4, , (A), , 19, 8, , (B), , 8, 19, , (C), , 19, 12, , 1,  –1 3, + tan –1  = tan, Solution (A) is the correct answer. tan  cos, 5, 4, , , 1,  –1 4, + tan –1 ,  tan, 3, 4, , , 20/04/2018

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34, , MATHEMATICS, ,  4 1 ,  3+4 ,  19  19, = tan tan –1   = ., , –1 , = tan tan , 4 1,  8 8, 1− × , 3 4, , , Example 38 The value of the expression sin [cot–1 (cos (tan–1 1))] is, (A) 0, , (B) 1, , (C), , 1, 3, , (D), , 2, ., 3, , Solution (D) is the correct answer., sin [cot–1 (cos, ,  –1 2 , 2, 1, π, =, )] = sin [cot–1, ]= sin sin, 3, 3, 2, 4, , ,  1 , Example 39 The equation tan–1x – cot–1x = tan–1 ,  has,  3, (A) no solution, (B) unique solution, (C) infinite number of solutions, (D) two solutions, , Solution (B) is the correct answer. We have, tan–1x – cot–1x =, , π, π, and tan–1x + cot–1x =, 6, 2, , Adding them, we get 2tan–1x =, , 2π, 3, , ⇒ tan–1x =, , π, i.e., x = 3 ., 3, , Example 40 If α ≤ 2 sin–1x + cos–1x ≤β , then, (A) α =, , −π, π, , β=, 2, 2, , (B) α = 0, β = π, , (C) α =, , −π, 3π, , β=, 2, 2, , (D) α = 0, β = 2π, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , Solution (B) is the correct answer. We have, , ⇒, , −π π, π, π, π, +, +, ≤ sin–1x +, ≤, 2, 2, 2, 2, 2, –1, –1, –1, 0 ≤ sin x + (sin x + cos x) ≤ π, , ⇒, , 0 ≤ 2sin–1x + cos–1x ≤ π, , ⇒, , 35, , −π, π, ≤ sin–1 x ≤, 2, 2, , Example 41 The value of tan2 (sec–12) + cot2 (cosec–13) is, (A) 5, , (B) 11, , (C) 13, , (D) 15, , Solution (B) is the correct answer., tan2 (sec–12) + cot2 (cosec–13) = sec2 (sec–12) – 1 + cosec2 (cosec–13) – 1, = 22 × 1 + 32 – 2 = 11., 2.3 EXERCISE, Short Answer (S.A.), , , , , 5 , 13 , –1 ,  + cos  cos, ., 6 , 6 , , , 1., , –1, Find the value of tan  tan, , 2., , –1, Evaluate cos cos, , 3., , Prove that cot, , 4., , Find the value of tan, , 5., , 2 , , Find the value of tan–1  tan  ., 3 , , , 6., , Show that 2tan–1 (–3) =, , π, 4, , – 3, π, + ., 2, 6, , – 2 cot –1 3 = 7 ., –1, , –, , 1, + cot –1, 3, , 1, –π, + tan –1 sin, 2, 3, , ., , –π, –1  –4 , + tan   ., 2,  3 , , 20/04/2018

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36, , 7., , MATHEMATICS, , Find the real solutions of the equation, , tan –1, , x ( x + 1) + sin –1 x 2 + x + 1 =, , , ., 2, , 8., , , –1 1 , –1, Find the value of the expression sin  2 tan,  + cos tan 2 2 ., 3, , , , 9., , If 2 tan–1 (cos θ) = tan–1 (2 cosec θ), then show that θ =, , (, , ), , , ,, 4, , where n is any integer., 10., , , , –1 1 , –1 1 , Show that cos  2 tan,  = sin  4 tan, ., 7, 3, , , , 11., , 3, , Solve the following equation cos tan –1 x = sin  cot –1  ., 4, , , (, , ), , Long Answer (L.A.), –1, , 1 + x 2 + 1– x 2, , =, , π, , 1, + cos –1 x 2, 4 2, , 12., , Prove that tan, , 13., , –1, Find the simplified form of cos, , 14., , –1, Prove that sin, , 8, 3, 77, + sin –1 = sin –1, ., 17, 5, 85, , 15., , –1, Show that sin, , 5, 3, 63, + cos –1 = tan –1, ., 13, 5, 16, , 16., , –1, Prove that tan, , 17., , –1, Find the value of 4 tan, , 1 + x – 1– x, 2, , 2, , 3, 4, –3π π, cos x + sin x , where x ∈, ,, ., 5, 5, 4 4, , 1, 2, 1, + tan –1 = sin −1, ., 4, 9, 5, 1, 1, – tan –1, ., 5, 239, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , 37, , 4+ 7, 1 –1 3, 4– 7, and justify why the other value, =, sin, 3, 2, 4, 3, , 18., , Show that tan, , 19., , is ignored?, If a1, a2, a3,...,an is an arithmetic progression with common difference d, then, evaluate the following expression., , , ,  d , d, d, –1 , –1 , tan  tan –1 ,  + tan ,  + tan ,  1 + a1 a2 ,  1 + a2 a3 ,  1 + a3 a4, , , , d, –1 ,  + ... + tan , ,  1 + an –1 an, , ,  ., , , Objective Type Questions, Choose the correct answers from the given four options in each of the Exercises from, 20 to 37 (M.C.Q.)., 20., , Which of the following is the principal value branch of cos–1x?, (A), , 21., ,  –  ,  2 , 2 , , (B), , , (0, π) –  , 2, Which of the following is the principal value branch of cosec–1x?, , (C), , [0, π], , (D), , (A), ,  –  , , , ,  2 2, , (B), , 23., , , [0, π] –  , 2, ,  –  , (D),  2 , 2 , If 3tan–1 x + cot–1 x = π, then x equals, ,  –  ,  2 , 2  – {0}, , (A), , –1, , (C), 22., , (0, π), , 0, , (B), , The value of sin–1, , (A), , 3, 5, , (B), , 1, , cos, , (C), , 33π, 5, , is, , –7π, 5, , (C), , π, 10, , (D), , 1, ., 2, , (D), , –π, 10, , 20/04/2018

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38, , MATHEMATICS, , 24., , The domain of the function cos–1 (2x – 1) is, (A), [0, 1], (B), [–1, 1], (C), ( –1, 1), (D), [0, π], , 25., , The domain of the function defined by f (x) = sin–1, (A), (C), , 26., , [1, 2], [0, 1], , , , , –1, If cos  sin, , 28., , –1, The value of cos cos, , π, 2, , , 6, , If tan–1 x + tan–1y =, , , 5, , (A), , 31., , 2, 5, , (C), , 3π, 2, (B), , If sin, , –1, , (B), , 0, , (D), , 1, , .96, , (D), , sin 1.5, , 5π, 2, , (D), , 7π, 2, , (D), , 1, , (D), , π, , is equal to, , 3π, 2, , (C), , The value of the expression 2 sec–1 2 + sin–1, , (A), 30., , (B), , The value of sin (2 tan–1 (.75)) is equal to, .75, (A), (B), 1.5, (C), , (A), , 29., , [–1, 1], none of these, , 2, , + cos –1 x  = 0 , then x is equal to, 5, , , 1, 5, , (A), 27., , (B), (D), , x –1 is, , 5, 6, , (C), , 1, 2, , is, , 7, 6, , 4, , then cot–1 x + cot–1 y equals, 5, (B), , 2, 5, , (C), , 3π, 5, , 2, 2a, 2x, –1 1– a, + cos, = tan –1, , where a, x ∈ ]0, 1, then, 2, 2, 1+ a, 1+ a, 1– x 2, , the value of x is, (A), , 0, , (B), , a, 2, , (C), , a, , (D), , 2a, 1– a 2, , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , 32., , The value of cot cos, , (A), , 33., , 25, 24, , –1, , 7, 25, (B), , The value of the expression tan, , 39, , is, , 25, 7, , 1, 2, cos –1, 2, 5, , (A), , 2+ 5, , (B), , 5–2, , (C), , 5+2, 2, , (D), , 5+ 2, , 24, 25, , (C), , (D), , 7, 24, , (D), , π, , is, , , θ, 1– cos θ ,  Hint :tan =, , 2, 1+ cos θ , , , 34., , If | x | ≤ 1, then 2 tan–1 x + sin–1, , (A), , 4 tan–1 x, , (B), , 2x, 1+ x 2, 0, , is equal to, , (C), , π, 2, , 36., , If cos–1 α + cos–1 β + cos–1 γ = 3π, then α (β + γ) + β (γ + α) + γ (α + β), equals, (A), 0, (B), 1, (C), 6, (D), 12, The, number, of, real, solutions, of, the, equation, , 37., , π , 1+ cos 2 x = 2 cos –1 (cos x )in  , π is, 2 , (A), 0, (B), 1, (C), –1, –1, If cos x > sin x, then, , 35., , (A), (C), , 1, < x≤1, 2, , −1≤ x <, , 1, 2, , (B), , 0≤x<, , (D), , x>0, , 2, , (D), , Infinite, , 1, 2, , 20/04/2018

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40, , MATHEMATICS, , Fill in the blanks in each of the Exercises 38 to 48., 38., ,  1, The principal value of cos–1  –  is__________.,  2, , 39., ,  3π , The value of sin–1  sin  is__________., 5 , , , 40., , If cos (tan–1 x + cot–1, , 41., , 1, The set of values of sec–1   is__________., 2, , 42., , The principal value of tan–1, , 43., 44., , 3 ) = 0, then value of x is__________., , 3 is__________., , 14π , , The value of cos–1  cos,  is__________., 3 , , The value of cos (sin–1 x + cos–1 x), |x| ≤ 1 is______ ., , 45., ,  sin –1 x + cos –1 x , 3,  ,when x =, The value of expression tan , is_________., 2, 2, , , , 46., , If y = 2 tan–1 x + sin–1, , 47., 48., , 2x, 1+ x 2, , for all x, then____< y <____., ,  x− y , The result tan–1x – tan–1y = tan–1  1+ xy  is true when value of xy is _____, , , –1, The value of cot (–x) for all x ∈ R in terms of cot–1x is _______., , State True or False for the statement in each of the Exercises 49 to 55., 49., 50., 51., 52., 53., , All trigonometric functions have inverse over their respective domains., The value of the expression (cos–1 x)2 is equal to sec2 x., The domain of trigonometric functions can be restricted to any one of their, branch (not necessarily principal value) in order to obtain their inverse functions., The least numerical value, either positive or negative of angle θ is called principal, value of the inverse trigonometric function., The graph of inverse trigonometric function can be obtained from the graph of, their corresponding trigonometric function by interchanging x and y axes., , 20/04/2018

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INVERSE TRIGONOMETRIC FUNCTIONS, , 41, , n π, > , n∈N , is valid is 5., π 4, , 54., , The minimum value of n for which tan–1, , 55., ,   –1 1  , π, The principal value of sin–1  cos  sin,   is ., 2 , 3,  , , 20/04/2018