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INVERSE TRIGONOMETRIC FUNCTIONS, , INVERSE TRIGONOMETRIC FUNCTIONS, 1. INVERSE OF SINE FUNCTION, , b. cos: R   1,1, , f  x   sin x, , a., , c. cos:...  , 0 or 0,   or  , 2 ...   1,1 is, both one-one and onto, , Y, , d. cos 1 :  1,1  0,   is called inverse function, , –3/2 –/2 /2 3/2, sin=X, , –, 2, O, –2, , , , y=1, X, y = –1, , value branch of cos, , 1, , f  x   cos1 x, , y, , Y, , , , b. sin: R   1,1, , , 2, ,  3         3 , ,, or, , or, ,, ..., c. sin:... , 2   2 2   2 2 ,  2,   1,1, is both one-one and onto, , sin, , , , 1, , tan=X, , –, , , 2, , cos=X, , CHETHAN M G, , O, , , , 3/2, , X, , , , , : nI  R, 2, , , , 3, , , , , , , ,  ,    3 , , Y, , Y, , /2, , –/2, , b. tan : R   2n  1, , , , , , –, , Y, , 2. INVERSE OF COS FUNCTION, , O, –3/2 – –/2 /2, , –3/2, , x, ,    , , , e. sin 1  sin x   x , x  ,  2 2, f. sin sin 1 x  x , x   1, 1, f  x   cos x, , f  x   tan x, , Y, , (1,0), , a., , , , y, , , 2, , x, , 3. INVERSE OF TAN FUNCTION, , a., , (–1,0), , , , (1,0), , f. cos cos1 x  x , x   1,1, , function of sin function, where , ,  is range,  2 2, or principal value branch of, , –1, , e. cos1  cos x   x , x  0,  , ,    , ,  is called inverse, d. sin 1 :  1,1  ,  2 2,    , , f  x   sin 1 x, , , , of cos function, where 0,  is range or principal, , y=1, X, 3/2 y = –1, , c. tan :... , ,,  or  ,  or  ,  ...  R is both, 2   2 2 2 2 ,  2, one-one and onto,    , d. tan 1 : R  , ,  is called inverse function of,  2 2, tan function, where     is range or principal, , , ,  2 2, , value branch of tan 1, , f  x   tan 1 x, [email protected], Phone number-8105418762 / 7019881906, , 1

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , y, , Y, , /2, x, , cosec=X, (–/2,–1), ,   , e. tan 1  tan x   x , x  , , ,  2 2, , 4. INVERSE OF COT FUNCTION, , f  x   cot x, Y, , –3/2, , –, , /2, , –/2, O, , 3/2, , , y = –1, , Y, , , , cot=X, –2, , (3/2,–1), , b. cos ec : R  n : n  I   R   1,1, , f. tan tan 1 x  x , x  R, , a., , X, , O, , –/2, , , , y=1, , (/2, 1), , (–3/2,1), , 2 X, , c. cos ec :...  3 ,      or   ,    0 or   , 3    ...,  2 2 ,  2 2, 2 2 , , , , , , ,  R   1,1, is both one-one and onto,    , d. cos ec 1 : R   1,1  , ,   0 is called,  2 2, inverse function of cosec function,,   , where , ,   0 is range or principal value,  2 2, branch of cos ec 1, , f  x   cos ec 1 x, , y, , Y, , b. cot: R  n : n  I   R, , y = /2, , c. cot :...   , 0  or  0,   or  , 2  ...  R, is both one-one and onto, d. cot 1 : R   0,   is called inverse function of cot, , (–1,0), , function, where  0,   is range or principal value, , O, , y = –/2, , branch of cot 1, f  x   cot 1 x, , y=, (0, /2), , 6. INVERSE OF SEC FUNCTION, , a., , x, , e. cot, ,    , ,   0,  2 2, f. cos ec  cos ec 1 x   x , x  R   1,1, e. cos ec 1  cos ec x   x , x  , , y, , 1, , x, , (1,0), , f  x   sec x, ,  cot x   x , x   0,  , , f. cot  cot 1 x   x , x  R, , sec=X, , (–2,1), –3/2, , Y, , (2,1), , (0,1), /2, , –/2, , –, –2, (––1), , 2, (–1), , 5. INVERSE OF COSEC FUNCTION, , 5/2, , 3/2, , O, , y=1, , y = –1, , f  x   cos ecx, , a., , Y, , 2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , X, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , , , , b. sec: R   2n  1, , , , : n  I   R   1,1, 2, , , c. sec :...  , 0     or 0,       or  , 2    3  ...,  2 , 2, 2,  R   1,1, , 6., , if    0,   ,  , ,  sec    ;, cot 1 cot    ;, , 5. sec, , 1, , if   0,  , , , 2, , PROPERTY –II, , , , if x   1,1, cos  cos x   x;, if x   1,1, tan  tan x   x;, if x  R, cos ec  cos ec x   x; if x   , 1  1,  , sec  sec x   x; if x   , 1  1,  , cot  cot x   x;, if x  R, , 1. sin sin 1 x  x;, , is both one-one and onto, ,  , 2, , 2., , d. sec 1 : R   1,1  0,      is called inverse, , 3., ,  , function of sec function, where  0,      is range, 2, 1, or principal value branch, of s ec, y, , f  x   sec1 x, , 4., 5., 6., , 1, , 1, , 1, , 1, , 1, , (0,), PROPERTY –III, , 1. sin 1  x    sin 1 x;, (0,/2), , (–1,0), , O, , (–1,0), , x, , 3., 4., , , e. sec 1  sec x   x , x  0,     , 2, 1, f. sec sec x  x , x  R   1,1, , , , , , 5., , DOMAIN AND RANGE OF INVERSE TRIGONOMETRIC, FUNCTION, , Function, , Domain, , Range, , sin 1 x, ,  1,1, , cos1 x, ,  1,1, ,   ,   2 , 2 , , tan 1 x, , R, , cot 1 x, , R, , sec x, , R   1,1, , cos ec 1 x, , R   1,1, , 1, , 2. cos cos    ;, 1, , 3. tan 1 tan     ;, , 0,  , , 4. cos ec 1 cos ec    ;, CHETHAN M G, , if x   1,1, if x   1,1, if x  R, , if x   , 1  1,  , 6. cot 1   x     cot 1 x;, , if x  R, , 1. sin 1  1   cos ec 1 x, if x   , 1  1,  , x, 2. cos 1  1   sec 1 x,, if x   , 1  1,  , x, ,   ,  , ,  2 2, , 0,  , , if x  0, , cot x;, 1, 3. tan 1    , 1, , 0,     , 2,   ,  2 , 2   0, ,  x      cot x;, , PROPERTY –V, , 1. sin 1 x  cos 1 x , ,   , if    , ,  2 2, , if   0,  , ,   , if     , ,  2 2,   , , 1, , PROPERTY- IV, , PROPERTIES OF INVERSE TRIGONOMETRIC, FUNCTION, PROPERTY -I, , 1. sin 1 sin     ;, ,   x     cos x;, tan 1   x    tan 1 x;, cos ec 1   x     cos ec 1 x;, if x   , 1  1,  , sec1   x     sec1 x;, , 2. cos, , 1, , 1, 1, 2. tan x  cot x , , , 2, , , , 1, 1, 3. sec x  cos ec x , , 2, , 1, , if x  0, , ; if x   1,1, , ;, , if x  R, , , ; if x   , 1  1,  , 2, , PROPERTY VI, , 1. sin 1 x  sin 1 y , , if    ,    0,  2 2, [email protected], Phone number-8105418762 / 7019881906, , 3

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, , CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , , ,  sin 1 x 1  y 2  y 1  x 2 ;, if  1  x, y  1& x 2  y 2  1 or, , , if xy  0 & x 2  y 2  1, , , 1, 2, 2, 2, 2,    sin x 1  y  y 1  x ; if 0  x, y  1& x  y  1, ,   sin 1 x 1  y 2  y 1  x 2 ;, if  1  x, y  0 & x 2  y 2  1, , , , , , , 1, , 2. sin, , , , , , x  sin, , , , 1, , , , 1. sin x  cos, , , ,  sin 1 x 1  y 2  y 1  x 2 ;, if  1  x, y  1& x 2  y 2  1 or, , , if xy  0 & x 2  y 2  1, , , 1, 2, 2, 2, 2,    sin x 1  y  y 1  x ; if 0  x  1, 1  y  0 & x  y  1, ,   sin 1 x 1  y 2  y 1  x 2 ; if  1  x  0,0  y  1& x 2  y 2  1, , , , , , , , PROPERTY VII, , 1. cos 1 x  cos 1 y , , , , , ,  cos 1 xy  1  x 2 1  y 2 ; if  1  x, y  1& x  y  0, , , 2  cos 1 xy  1  x 2 1  y 2 ; if  1  x, y  1& x  y  0, , 2. cos 1 x  cos 1 y , , , , , , , , , , , cos 1 xy  1  x 2 1  y 2 ; if  1  x, y  1& x  y, , ,  cos 1 xy  1  x 2 1  y 2 ; if  1  y  0,0  x  1& x  y, , , , , , , PROPERTY VIII, , 1. tan, , 1, , x  tan 1 y , , ,  x y , tan 1 , ,  ; if xy  1,  1  xy , , , 1  x  y ,   tan ,  ; if x  0, y  0 & xy  1,  1  xy , , ,  x y ,   tan 1 ,  ; if x  0, y  0 & xy  1, ,  1  xy , , 2. tan 1 x  tan 1 y , , ,  x y , tan 1 , ,  ; if xy  1, 1, , xy, , , , , 1  x  y ,   tan ,  ; if x  0, y  0 & xy  1,  1  xy , , ,  x y ,   tan 1 ,  ; if x  0, y  0 & xy  1, ,  1  xy , 4, , PROPERTY IX, 1, , y, , , , , xy  1, yx, xy  1, 4. cot 1 x  cot 1 y  cot 1, yx, 3. cot 1 x  cot 1 y  cot 1, , 1, , 1  x  tan, 2, , x, , 1, , 1  x2, ,  cot, , 1, , 1  x2, x, ,  1 , 1,  sec 1 ,  cos ec 1  , , 2, x,  1 x , 2. cos 1 x  sin 1 1  x 2  tan 1, , 1  x2, x,  cot 1, x, 1  x2, ,  1 , 1,  sec 1    cos ec 1 , , 2, x,  1 x ,  x ,  1 , 1  1 , 3. tan 1 x  sin 1 ,  cos 1 , ,   cot  , 2, 2,  x,  1 x ,  1 x ,  1  x2 ,  cos ec 1 ,   sec 1 1  x 2,  x , , , , , , PROPERTY X, , , , , , , , 1, 1, , 1, 2, x,  sin 2 x 1  x ; if , 2, 2, , 1, , 1. 2sin 1 x     sin 1 2 x 1  x 2 ; if,  x 1, 2, , , 1, 1, 2,   sin 2 x 1  x ; if  1  x  , 2, , 1, 2,  cos  2 x  1 ; if 0  x  1, 2. 2 cos 1 x  , , 1, 2, , 2  cos  2 x  1 ; if  1  x  0, , , , , , , , , , , 1  2 x ,  tan  1  x 2  ; if  1  x  1, , , , ,  2x , 3. 2 tan 1 x     tan 1 , ; if x  1, 2 ,  1 x , , , 1  2 x , ; if x  1,   tan , 2 ,  1 x , , , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , PROPERTY XI, , 1, 1, , 1, 3,  sin 3 x  4 x ; if  2  x  2, , 1, 1. 3sin 1 x     sin 1 3x  4 x 3 ; if  x  1, 2, , 1, , 1, 3,   sin 3x  4 x ; if  1  x   2, , 1, , 1, 3,  cos 4 x  3x ; if 2  x  1, , 1, 1, 1, 2. 3cos x   2  cos 1 4 x3  3x ; if   x , 2, 2, , 1, , 1, 3, 2  cos 4 x  3x ; if  1  x   2, , , , , , , , , , , , , , , , , , , , , , ,  1  3 x  x , 1, 1, ; if , x,  tan , 2 , 3, 3,  1  3x , , ,  3x  x3 , 1, , ; if x , 3. 3 tan 1 x     tan 1 , 2 , 3,  1  3x , , , 3, , ,    tan 1  3 x  x  ; if x   1, 2, , 3,  1  3x , 3, , PROPERTY XII, , 2., , 3., , 1, , If tan x  tan y  tan z , , then xy  yz  zx  1, , 2, , ,, , If tan 1 x  tan 1 y  tan 1 z   ,, then x  y  z  xyz, CHETHAN M G, , If cos 1 x  cos 1 y  cos 1 z  3 , then, xy  yz  zx  3 & x  y  z  1, , 7., , If cos 1 x  cos 1 y  cos 1 z   , then, , x 2  y 2  z 2  2 xyz  1, 8., , If sin 1 x  sin 1 y  sin 1 z , , then xy  yz  zx  3, , 3, ,, 2, , If sin 1 x  sin 1 y   , then, cos 1 x  cos 1 y    , 1, 1, 10. If cos x  cos y   , then, sin 1 x  sin 1 y    , , 11. If tan 1 x  tan 1 y  , then xy  1, 2, , 1, 1, 2, 2, 9., , 12. If cos A  cos B , , 2, , , then A  B  1, , , 2, , , then xy  1, , x, y,  cos 1   ,, a, b, , x 2 2 xy, y2, , cos, , ,  sin 2 , 2, 2, a, ab, b, , , , , x,  sin 1, , 2, 2, a,  a x ,  3a 2 x  x3 , x,  3 tan 1, 16. tan 1 , 2, 2 , a,  a(a  3x ) , x, , 15. tan 1 , ,  x  y  z  xyz , 1. tan x  tan y  tan z  tan , ,  1  xy  yz  zx , , 1, 1, 1, 1, , ,, , 6., , then, , SHORT CUT METHODS, , 2, , x 1  x2  y 1  y 2  z 1  z 2  2 xyz, , 14. If cos 1, , 2, , 1  1  x , ; if 0  x  ,  cos , 2 , ,  1 x , 1, 2. 2 tan x  , 2,  cos 1  1  x  ; if    x  0, , 2 , ,  1 x , , , 1, , 5., , then x 2  y 2  z 2  2 xyz  1, If sin 1 x  sin 1 y  sin 1 z   , then, , 13. If cot 1 x  cot 1 y , ,  1  2 x ,  sin  1  x 2  ; if  1  x  1, , , , ,  2x , 1. 2 tan 1 x     sin 1 , ; if x  1, 2 ,  1 x , , , 1  2 x , ; if x  1,   sin , 2 ,  1 x , , , 1, , If sin 1 x  sin 1 y  sin 1 z , , , , , , , , 4., ,  1  x2  1  x2   1, 1 2, 17. tan 1 ,    cos x, 2, 2,  1  x  1  x  4 2, , 1 x 1,  cos 1 x, 1 x 2, 1, 1 y, , 19. If cos x  cos, a, b, 18. tan 1, , then, , x 2 2 xy, y2, , cos, , ,  sin 2 , a 2 ab, b2, ,  a   sin, , 1, 20. If sin x, , 1, , yb  , , [email protected], Phone number-8105418762 / 7019881906, , 5

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, 2, , (a) Does not exist, (c) 1, , 2, , x, 2 xy, , cos   y 2  sin 2 , 2, b, a, ab, 1, 1, 21. asin x  b cos x  c then,  ab  c  a  b , asin 1 x  b cos 1 x , ab, , then, , TYPE:1 FIND THE VALUES, , [KCET-2020-1m] [Only one option correct], , Solution:(a) Does not exist because, 1, , , (a), 2, , 5, (b), 3, , 10, (c), 3, , (a) , , (d) 0, , 5, cos 5  , Solution:(a) cos 1 cos   sin 1 , , 3 , ,  3  2, , , 1, 1,  sin x  cos x  , 2, , , 5, , 3. Value of cos 1  cos, 3, , , 5 , , 1 ,   sin  sin,  is, 3 , , , , 2, 10, (a)0, (b), (c), (d), 3, 2, 3, 5, , 5, , , , , , 1, Solution:(a) cos 1  cos,   sin  sin, , 3, 3 , , , , ,  ,  ,  ,  ,  cos 1 cos  2     sin 1 sin  2   , 3 , 3 ,  ,  ,  ,   0, 3 3, 2, 4. If sin 1 x  sin 1 y , , then, 3, cos 1 x  cos 1 y , 2, , , (a), (b), (c), (d) , 3, 3, 6, 2, Solution:(b) sin 1 x  sin 1 y , 3, , , 2, ,  cos 1 x   cos 1 y , 2, 2, 3, ,  cos 1 x  cos 1 y  ., 3, , , , 5. The value of cos  sin 1  cos 1  is, 3, 3, , 6, , , , 3, ,   1,1, ,  3,  is,  2 , , 6. The principal value of sin , , , LEVEL-I, , 1. sec2(tan–12) + cosec2(cot–13) =, a) 1, b) 5, c) 10, d) 15, Solution:(d) 1 + tan2(tan–12) + 1 + cot2(cot–13)= 1 + 22, + 1 + 32 = 15, 5 , 5 , 1 , 2. The value of cos 1  cos,   sin  cos,  is, 3 , 3 , , , , (b) 0, (d) 1, , 2, 3, , (b) , , , 3, , 4, 3, ,   , (c), , (d), , 5, 8, , , Solution:(b) sin 1 sin      , 3,   3 , , , , 1,    sin x  , 2, 2, , 7. If   sin 1[sin ( 600)] , then one of the possible, value of  is, , 2, 2, (d) , 3, 3, 1, o, Solution:(a)   sin [sin(600 )], = sin 1[  sin(360  240)],   sin 1[ sin 240o ]  sin 1[ sin(180o  60o )] , (a), , , 3, , (b), , , 2, , (c), ,     ,   sin 1 sin 60o  sin 1 sin     ., 3, 3, , , , , 8., , 1, , , , sec [sec(30 )] , o, o, (a) – 60, (b) 30, o, o, (c) 30, (d) 150, o, , Solution:(c) sec1[sec(30o )]  sec1 (sec30o )  30o, 12, 4, 63, 9. sin 1  cos 1  tan 1, , 13, 5, 16, (a)0, , (b), , , 2, , (c) , , (d), , 3, 2, , Solution:(c) tan 1 12  tan 1 3  tan 1 63, , 5, 4, 16, 48, , 15, 63, ( xy  1 ),    tan 1,  tan 1, 20  36, 16, 63, 63,    tan 1,  tan 1,  ., 16, 16, , , ,  x y , tan 1 x  tan 1 y    tan 1 ,  ; if x  0, y  0& xy  1,  1  xy , , , , 10. sin 1, (a), , , 6, , 1,  cot 1 3 is equal to, 5, (b), , , 4, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , (c), , , 3, , (d), , , 2, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , Solution:(b) sin 1, , 1, , , , , , , 1,  1 , 1, 5   cot 1 3,  cot 1 3  cot 1 ,  1 , 5, , 5 , ,  ., 1, 1  2  3  1 , 1, ,  cot (2)  cot (3)  cot ,   cot (1) , 4,  3 2 , xy  1, 1  x2 , cot 1 x  cot 1 y  cot 1, & sin 1 x  cot 1, , yx, x , , 3, , then, 2, 9, the value of x100  y100  z100  101 101 101, x y z, is equal to, (a)0, (b)3, (c)– 3, (d)9, 1, Solution:(a) As we know sin x cannot be greater, , , 1, 1, 1, than ,So sin x  sin y  sin z , 11. If sin 1 x  sin 1 y  sin 1 z , , 2, , 2, , Therefore x  y  z  1, Putting these values in the expression, we get, 111, , 12. cot 1, , 9,  0., 111, , (a)0, (b)1, (c) cot 1 x  cot 1 y  cot 1 z, (d)None of these, xy  1, yz  1, zx  1, Solution:(a) cot 1,  cot 1,  cot 1, x y, yz, zx, ,  cot 1 y  cot 1 x  cot 1 z  cot 1 y  cot 1 x  cot 1 z  0 ., 13. If cos x  cos y   then sin 1 x  sin 1 y is, 4, , (a)0, (b), (c) 3, (d) , 4, 4, 1, , Solution:(c) If cos 1 x  cos 1 y   then, , sin 1 x  sin 1 y    , If sin 1 x  sin 1 y   then cos 1 x  cos 1 y    , , sin 1 x  sin 1 y        , 1, , 1, , 14. If tan 1  tan 2  tan, , 1, , 4, ,  3, , 4, , x   then the value of, , x, (a) 1, (b) 0, (c) 1, (d) 3, 1, 1, Solution:(d) If tan x  tan y  tan 1 z   then, , x  y  z  xyz, tan 1 1  tan 1 2  tan 1 x   then, 1 2  x  1 2  x, 3x  x  2x  x  3, CHETHAN M G, , (c) , , (d) 3 / 2, , x   2  x 1, x, , Solution:(b), , Therefore the principal value of sin 1 x , 16. sin 1 x  sin 1, , (b), , 3, 2, , (c), , , 2, , ., , 1, 1,  cos 1 x  cos 1 , x, x, , (a) , , , 2, , (d) None of these, , 1, 1,  cos 1 x  cos 1, x, x, , ,  sin 1 x  cos ec 1 x  cos 1 x  sec 1 x    , 2 2, 1, 17. The value of sin  2sin 0.8 , is equal to, 1, 1, Solution:(a) k  sin x  sin, , (a) sin1.20, , xy  1, yz  1, zx  1,  cot 1,  cot 1, , x y, yz, zx, , 1, , 1,  2 , the principal value of sin 1 x is, x, (a)  / 4, (b)  / 2, , 15. If x , , (b) 0.96 (c) 0.48, , (d) sin1.60, , [KCET-14-1m] [Only one option correct], 1, Solution:(c) Let sin 0.8    sin   0.8  sin  , , 5, 4, ,  cos  , , , 3, , , , 4, 5, , 3, 5, , , , sin 2sin 1 0.8  sin 2,  4  3  24,  2sin  cos   2      0.96,  5  5  25, 1, 1, 1, 18. If   sin x  cos x  tan x , x  0 , then the, smallest interval in which  lies is given by, , 3, (a)   , (b) 0    , 2, 4, , , , (c)     0, (d)   , 4, 2, 4, Solution:(b), ,    sin1 x  cos1 x  tan1 x = 2  tan 1 x, , We know , , , 2, ,  tan 1 x , , , , , ,   tan 1 x  , 2, 2, , , , 0, , , 2, , , 2, ,  tan 1 x   ., , [email protected], Phone number-8105418762 / 7019881906, , 7

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, 1, , 19. If in a triangle ABC, A  tan 2 and B  tan 1 3,, then angle C is equal to, (a)  / 2, (b)  / 3, (c)  / 4, , (d) None of these, , 1, 1, Solution:(c)Given that A  tan 2, B  tan 3, , We know that A  B  C  , , 53 , , Solution:(d) sin 1  cos, , 5 , ,  , 3  ,  sin 1  cos 10     cos  2n     cos , 5 ,  ,   3  ,  sin 1  cos     sin       cos  , 2, ,   5  , ,    3  ,  sin 1  sin    ,   2 5 ,   5  6  ,  sin 1  sin , ,   10  , ,  tan 1 2  tan 1 3  C  ,  23 ,  tan 1 ,   C  , 1  2  3 , ,  tan 1 (1)  C  , 3, , ,  C    C  ., 4, 4, a, a,  1,  1, 20. tan   cos 1   tan   cos 1  equal, b, b, 4 2, 4 2, to, 2b, a, b, 2a, (a), (b), (c), (d), a, b, a, b, , a, a,    cos  , b, b, a, a,  1,  1, tan   cos 1   tan   cos 1  =, b, b, 4 2, 4 2, 1 t 1 t, , , , where t  tan, 1 t 1 t, 2, , 1, Solution:(b) Let cos, , 2, , 2, , (b), , , 2, , 5  2,  cos 1 x  sin 1 y, 5, 3,  cos 1 x  sin 1 y, 5, 3, cos 1 x  sin 1 y , 5, 1, 1, 4, , 24. If tan x  tan y , then cot 1 x  cot 1 y is, 5, equal to, , 2, , (c)0, , (d)None of these, ,  xz ,  xy ,  yz , Solution:(b) tan 1    tan 1    tan 1  ,  zr ,  xr ,  yr , ,  xy yz xz xyz ,     3 , , xr yr r , 1  zr,  tan,  tan 1   ., 2, 2, 2, , 2, x y z  ,  1 ,  , 2, r, ,  , , 53 , , 22. The value of, sin 1  cos,  is, 5 , , (a) 3, 5, , (b) 3, 5, , (d) , , 10, , 10, , 8, , (a) 2, , (b) , , (c) 3, , 5, , 5, , 5, , (d) , , [KCET-17-1m] [Only one option correct], , Solution:(b) tan 1 x  tan 1 y  4, , , , , (c) , , [KCET-16-1m] [Only one option correct], , , ,  sin 1 y  2 ,, 5, 2,  cos 1 x  sin 1 y, ,  cos 1 x , ,   2 5, ,  zx ,  yz ,  xy , tan 1    tan 1    tan   ,  zr ,  xr ,  yr , , (a) , , Solution:(b) sin 1 x  cos 1 y  2 ,, 5, , 2, , 21. If x  y  z  r , then, 2, , [KCET-18-1m] [Only one option correct], , , , (1  t 2 ), 2, 2b, ., , , 2, 1 t, cos , a, 2, ,          ,  sin 1  sin ,  , ,  ,   10   10  2 2 , 23. If sin 1 x  cos 1 y  2 , then cos 1 x  sin 1 y, 5, is, 2, 3, 4, 3, (a), (b), (c), (d), 5, 5, 5, 10, , 2, , 2, ,  cot 1 x ,  cot 1 x , , , , 2, , 2, , 5, ,  cot 1 y  4, ,  cot 1 y   4, , , , 5, 5, , , , cot 1 x  cot 1 y   4  , 5 2 2, 4  5, cot 1 x  cot 1 y   4   , 5, 5, cot 1 x  cot 1 y , , , , 5, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , 1, 2, , 25. The value of the expression tan  cos 1, (a) 2  5, , (c), , (b) 5  2, , 52, 2, , ,  x, , Solution:(c) tan sin 1 , , 2, , , , Put x  sin , ,  sin , 1  sin 2 ,  tan sin 1 , , 2, ,  2, , [KCET-18-1m] [Only one option correct], , 2, , 5, , 2,  cos   1, 5, 2, , 4,  2 , & sin   1  cos   1  ,   1 5,  5, 2, , 4, 54, 1, sin   1  , ,   2, 5, 5, 5, , 1,  , 1 2 ,  tan  , (ii) tan  cos, , 5, 2, 2,  ,  ,  , , sin   2sin   cos  , 2 , 2,  2   sin , , 1  cos,  ,  , cos  , 2 cos2  , 2, 2, 1, 5 2, 5  1, , , 2, 52, 52, 5 2, 1, 5, , , , , , 5 2, ,  , , , , , , 5 2,  5 2, 5  4, , , ,  5  22 , , , , , 26. The domain of the function defined by, f  x   cos 1 x  1 is, 2, , (a)  0,1, , (c)  0, 2, , (b) 1, 2, , (d)  1,1, , [KCET-2020-1m] [Only one option correct], , Solution:(b) f  x   cos 1 x  1, , W.k.t 1  y  1 & 0  cos 1  y   , ,  1  x  1  1, ,  0  x 1  1,  0  x 1  1, 1  x  2,  x  1, 2  Domain, 1, 27. Given 0  x  , then the value of, 2, CHETHAN M G, , , , , 1  x2 ,  x, , 1, tan sin 1 , ,   sin x  is, 2 , , , ,  2, , 1, (a) 3, (b), (c) 1, (d) 1, 3, [KCET-14-1m] [Only one option correct], , (d) 5  2, , Solution:(b) (i)Put cos 1, , 2 ,  is, 5, , , , 1  x2, 2, , , , , 1, , , sin, x, , , , , , , , 1,   sin  sin   , , , , , ,  tan sin 1 cos 450 sin   sin 450 cos    ,  tan sin 1 sin 450     , , , 0, 0,  tan 45       tan 45   1, ,  , , , , 28. The range of sec 1 x is,  , (a)  0,     , (b)    ,  ,  2 2, 2, , , (c)    ,  , (d)  0,  ,  2 2, [KCET-17-1m] [Only one option correct], [NCERT-DIRECT QUESTION], ,  , 2, 2 2, 1  1 , 29. The value of sin 1 , is equal to,  3   sin  3 , , , , , , 2, (a), (b), (c), (d), 3, 6, 2, 4, Solution:(a)  0,     , , [KCET-15-1m] [Only one option correct], , Solution:(b), , 2 2, 1, 1, sin 1         sin   cos  , 3, 3, 3, 2 2, , , , , 1  1 , 1 2 2, 1 2 2, sin 1 ,   sin    sin ,   cos , , 3,  3 ,  3 ,  3 , 2 2, , 1  1 , 3, sin 1 ,   sin   , 1, 3 2,  3 , , 3, 3, , , 2 2, 30. cos  2 sin 1  cos 1  , 4, 4, , 3, (a) does not exist, (b), 5, 3, 3, (c), (d), 4, 4, [KCET-19-1m] [Only one option correct], [email protected], Phone number-8105418762 / 7019881906, , 9

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 3, 3, , Solution:(d) cos  2sin 1  cos 1 , 4, 4, , 3, 3, 3, ,  cos sin 1  sin 1  cos 1 , 4, 4, 4, , 3 , 3  3, , ,  cos sin 1     sin  sin 1  , 4 2, 4 4, , , 31. If a , , , , 2, , 1, , 1, ,  2 tan x  3cot x  b .Then, , ' a '& ' b ' are respectively,  , (a), &, 2, 2, , , , (c), , 2, , (d) 0 & , , & 2, , Solution:(c) 0  cot 1 x  ,     cot 1 x  2,   2 tan 1 x  2 cot 1 x  cot 1 x  2, ,   2 tan 1 x  3cot 1 x  2  1, , Given, , , , 2, ,  2 tan 1 x  3cot 1 x  b   2 , , From 1&2  a , , , 2, ,   a , , 32. If x  n , x   2n  1, , , 2, , , 2, , & b  2, , , n  Z then, ,  sin  cos x   cos  sin x  , sin 1 , ,  tan 1  cot x   cot 1  tan x  , , , 1, , a), , , , 1, , b), , , , c), , , , d), , , 2, , 6, 4,  sin 1 (cos x)  cos 1 (sin x) , Solution:(d) sin 1 ,  tan 1 (cot x)  cot 1  tan x  , , , 3, ,  1   , , ,  , 1 ,  sin sin   x    cos cos   x   , ,  ,  2,  2,  sin 1 ,  1   , , ,  , 1 ,  tan  tan  2  x    cot cot  2  x   , ,  ,  ,  , , , , ,  x  x, , 1 2, 2,  sin , , , ,   x  x, 2, 2, , , ,  sin 1 (1) , 2, , 10, , 2, , , , x x, 3, , A Bx  C, , then, x x2  1, , 1, 1, cos ec 1    cot 1    sec 1  c  ,  A, B, a) 5, b) 0, c) , d) , 2, 6, 6, , [KCET-2015-1M] [One option correct type], ,  x  1, , Solution:(a), , 2, , , , A Bx  C,  2, x, x 1, , x x, 2,  x  1  A  Bx  C, x2  1, x  x 2  1 x, , (b) 0&2, , [KCET-19-1m] [Only one option correct], , a, , 33. If, ,  x  1, , 3, ,  x  1, , 2, , , , , ,  A x 2  1  x  Bx  C , , Put x  0  1  A, Comparing c of x 2 1  A  B 1  1 B  B  0, Comparing co-eff of x  2  C, 1, 1, 1, cosec 1    cot 1    sec 1  ,  A, B, C, ,  5, 1 ,  cosec 1 1  cot 1     sec 1     0  , 3, 6, 2 2, , 34. If 3  i   a  ib c  id, , , , b, d, then tan 1    tan 1  , a, c, a)   2n , n  I, , 3, , c) n , , , 3, , b) n , , , 6, , d) 2n , , , nI, , Solution:(b) 3  i   a  ib  c  id , , , n I, , , 3, , , nI, , 3  i  ac  adi  ibc  bd, 3  i   ac  bd    ad  bc  i, , ac  bd  3 , ad  bc  1, b, d ,  bc  ad , tan 1    tan 1    tan 1 , , a, c,  ac  bd ,  1 ,  tan 1 , ,  3, ,  n , , , , , n I, 6,  1 , 1,  tan 1 ,      tan , 3,  3,  tan, , , , 6, ,  tan     n , , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , , 6, , MCP-MATHEMATICS

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1, 1 y,   ,then 4 x 2  4 xy cos   y 2 is, 35. If cos x  cos, 2, , equal to, 2, (a) 4sin , (c) 2sin 2, , 2, , y, , a, , b, ,   then, , x 2 2 xy, y2, , cos, , ,  sin 2 , 2, 2, a, ab, b, x, y, 1,  cos 1   ;, Here cos, 1, 2, 2, x 2 xy, y2,  , cos  ,  sin 2 , 1, 2, 4, 2, y, x 2  xy cos  ,  sin 2 , 4, 2, 4 x  4 xy cos   y 2  4sin 2  ., 36. If 2sin 1 x  3cos 1 x  4 then 2sin 1 x  3cos 1 x, equals, , 6  4, 4  6, 0, (b), (c) 3, 2 (d), 5, 5, 1, 1, Solution:(a) If a sin x  b cos x  c then,  ab  c(a  b), a sin 1 x  b cos 1 x , ab, put a  2, b  3, c  4 wages,  (2)(3)  4(2  3) 6  4, 2sin 1 x  3cos 1 x , , 23, 5, 6  4, , 5, (a), , 37. The domain of the function defined by, , f  x   cos, , 1, , (a)  0,1, , (c)  0, 2, , x  1 is, (b) 1, 2, , (d)  1,1, , [KCET-2020-1m] [Only one option correct], , Solution:(b) f  x   cos 1 x  1, , W.k.t 1  y  1 & 0  cos 1  y   , ,  1  x  1  1, ,  0  x 1  1, ,  0  x 1  1, 1  x  2, ,  x  1, 2  Domain, , CHETHAN M G, , 1. If a, b, c be positive real numbers and the, , a(a  b  c), b( a  b  c ),  tan 1, bc, ca, c( a  b  c),  tan 1, , then tan  is, ab, (a)0 (b)1, (c) a  b  c (d)None of these, , Value of   tan 1, , (b) 4sin , (d) 4, , 1, 1, Solution:(d) If cos x  cos, , INVERSE TRIGONOMETRIC FUNCTIONS, LEVEL-II, , Solution:(a), a(a  b  c), b( a  b  c ), c(a  b  c),   tan 1,  tan 1,  tan 1, bc, ca, ab, 2, Let s , , abc, abc, ,    tan 1 a 2 s 2  tan 1 b2 s 2  tan 1 c 2 s 2, ,   tan 1 (as)  tan 1 (bs)  tan 1 (cs),  as  bs  cs  abcs 3 , 2, 2, 2, 1  abs  bcs  cas , ,   tan 1 , ,  (a  b  c)  abcs 2 , tan   s , 0, 2, 1, , (, ab, , bc, , ca, ), s, , , 2, [ abcs  (a  b  c)], Trick: Since it is an identity so it will be true for, any value of a,b,c. Let a  b  c  1 then, ,   tan 1 3  tan 1 3  tan 1 3   , tan   0 ., 1, 1, 1, 2. If sin x  sin y  sin z  3 and, 2, f (1)  2 , f ( x  y)  f  x  f  y  , x, y , , then, x, , f (1), ,  y f (2)  z f (3) , , x, , f (1), , x yz,  y f (2)  z f (3), , is, a)0, Solution:(c), , (b)1, , c) 2, , sin 1 x  sin 1 y  sin 1 z , , d) 3, , 3,  x  y  z 1, 2, , xy  yz  zx  3, f ( x  y )  f ( x)  f ( y ), f 1  1  f (1) f (1)  f (2)  (2)(2)  4, , f  2  1  f (2)  f (1)   4  2   8  f (3)  8, ,  x f (1)  y f (2)  z f (3) , , x, , f (1), , x yz,  y f (2)  z f (3), , 111, 1  14  18, 3,  111,  3 1  2, 111,  12  14  18 , , 2, , [email protected], Phone number-8105418762 / 7019881906, , 11

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 2   cos, , 1, 3. If cos x, , 1, , y3   0, , CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, ,  19, ,  cot   tan 1  n  1  tan 1  n  ,  n 1, , , , , then, , 9 x 2  12 xy cos   4 y 2 is, , , , 1, 1, 1  x  y ,  tan x  tan y  tan , ,  1  xy  , , , (b) 36sin 2 , (d) 36 cos 2 , , (a) 36, (c) 36sin 2 , , y , Solution:(c) If cos x  cos, then, a, b, x 2 2 xy, y2, , cos, , ,  sin 2 , a 2 ab, b2, Here a  2 , b  3, x2, 2 xy, y2, , cos, , ,  sin 2 , 2, 2, 2 (2)(3), 3, 2, 2, x, xy, y,  cos  ,  sin 2 , 4 3, 9, 2, 9 x  12 xy cos   4 y 2  36sin 2 , 1, 1, 1, 4. If tan x  tan y  tan z   then the, 2, 2, 2, 2, value of  x  y  z    x  y  z 2  is equal to, 1, , (a) 1, , (b) 2, , 1, , (c) 3, , (d) 4, , 1, 1, 1, Solution:(b)If tan x  tan y  tan z  , , xy  yz  zx  1, ,  x  y  z, , 2, , 2 then, ,  x 2  y 2  z 2  2(xy yz 2 x), , , , , , (x  y z)2  x2  y 2  z 2  2(1)  2, ,  cot[tan 1  2   tan 1 1  tan 1  3  tan 1  2 ,  tan 1  4   tan 1  3  ....  tan 1  20   tan 1 19 ], , , , 22, (c) 19, 21, , 23, 21, (d), 19, , [JEE Main-10/1/2019-Shift-II-4M], [One option correct type], , n,  19, , , 1, Solution:(d) cot   cot 1   2 p  ,  n 1, ,  p 1  , ,  19, , n  n  1  ,  cot   cot 1 1  2, ,  n 1, 2  , , ,  19, ,  cot   cot 1  n  n  1  1 ,  n 1, ,  19, , , 1,  cot   tan 1 ,  n  n  1  1  ,  n1, , , , , 12, , , , , , cot y  cot x , , ,  , , cot  tan 1   cot  tan  20  , , cot tan 1  20  cot tan 1 1  1, 1, , 1, , ,  1 , cot  cot 1    cot cot 1 1  1,  20  ,  , ,  1 , cot cot 1 1  cot  cot 1   ,  20  , , 1, 1  1 1  20 21,  20, , , 1, 20, , 1, 19, 1, 20, 1  3 , 1  1 , 6. If   cos   ,   tan   , where of, 5,  3, , , , , , , , , , , , then    is equal to:, 2, 9 , 9 , 1 , 1 , (a) tan , (b) sin , , ,  5 10 ,  5 10 , 9 , 1 , 1  9 , (c) cos , (d) tan  , ,  14 ,  5 10 , , 0  ,  , , n,  19, , , 1, cot, 1, , 2 p   is, , , ,  n 1, ,  p 1  , , , (b) 22, , , ,  cot[tan 1  20   tan 1 1]  cot  x  y   cot x cot y  1 , , 5. The value of cot , (a) 23, , , , [JEE Main-8/4/2019-Shift-I-4M], [One option correct type], , Solution:(b), , 3, 5, , 1, (i)   cos    cos  , , , , 3, 4,  sin  , 5, 5, , 3, 4,5 are pythagoras triplets , , 1, 3, 3, , 1, (ii)   tan   ,  tan  , , & cos  , , 1, 1,  sin  , 3, 10, ,  1,3, 10 are pythagoras triplets , , , 10 , (iii) w.k.t. sin      sin  cos   cos  sin ,  4  3   3  1 ,  sin        ,    , ,  5   10   5   10 , , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , Since x  5 & x 2  1 are always positive so, , 3 4 1, 3 3,  sin     ,   ,  , 10  5 5 , 10  5 , 9,  9 ,       sin 1 ,  sin     , , 5 10,  5 10 , , x 5, x2  1, , (ii) So for domain, ,  12 , 1  3 ,   sin  ,  13 , 5, , 1, 7. The value of sin , , , , sin 1 x  sin 1 y  sin 1 x 1  y 2  y 1  x 2 ;, , , , , , 2, 2, 2, 2, if  1  x, y  1& x  y  1 or if xy  0 & x  y  1, , (a), , 17  1, 2, 17, (d), 2, , [JEE Main-02/09/2020-Shift-I-4M], [One option correct type], ,  x 5, , 2,  x 1 , , Solution:(c) (i) f  x   sin 1 , ,  1 , , x 5, x2  1, , 1, , CHETHAN M G, , , 2, , (b), , 5, 4, , (c), , 3, 2, , (d), , 7, 4, , [JEE Main-3/9/2020-Shift-I-4M], [One option correct type], ,  x 5, 8. The domain of the function f  x   sin 1  2 ,  x 1 , , (b), , 1  17, 2, , is equal to:, , 33,56, 65 are pythagoras triplets , , 17, 1, 2, 1  17, (c), 2, , , , , 1  17, 1  17  1  17 ,  x   , , ,    a , , 2, 2   2, , , 4, 5, 16 , , 9. 2   sin 1  sin 1  sin 1 , 5, 13, 65 , , ,  12  4  3  5  ,  sin 1       ,  13  5  5  13  ,  48 15 ,  33 ,  sin 1     sin 1  ,  65 65 ,  65 , ,  33  ,  56 ,   cos 1     sin 1  , 2,  65  2,  65 , , (a), , 2, , , 1  17 , 1  1  17 ,   x , x, , ,   0, , 2, 2, , , , 1  17, 1  17,  x,  rejected case  or x , 2, 2, ,  x, , is  , a    a,   . Then a is equal to:, , 2, , , , Solution:(a) sin 1 , , , , 1, , , , 2,  x  x  4  0  x  1  1  16  1  1  17 , , [JEE Main-12/4/2019-Shift-I-4M], [One option correct type], , , , x2  1, , 2, 2,  x  5  x  1  x  x 2 , x  , , , , 1  63 , (b)   sin  ,  65 , 1  33 , (d)   cos  ,  65 , ,  12 , 1  3 ,   sin  ,  13 , 5,  12, 9 3, 144 ,  sin 1 , 1,  1, , 25 5, 169 ,  13, , x 5, ,  x  5  x2  1, , is equal to:, , , 1  56 , (a)  sin  , 2,  65 , , 1  9 , (c)  cos  , 2,  65 , ,  0, x , , , , , 4, 5, 16 ,  sin 1  sin 1 , 5, 13, 65 , 4, 5, 16 , ,  2   tan 1  tan 1  tan 1 , 3, 12, 63 , , , 1, Solution:(c) 2   sin, , , ,  4,3,5 , 5,13,12 , 16,65,63 are pythagoras triplets , , , , 4 5 ,   , , 16 , 3 12 ,  2   tan 1 ,  tan 1 , 63 , 4 5 , , 1   , , ,  3 12 , , , ,  48  15 , 1 16 ,  2   tan 1 ,   tan, , 63 ,  36  20 , , 63, 16 , ,  2   tan 1  tan 1 , 16, 63 , , 16, 16 ,  3, ,  2   cot 1  tan 1   2  , 63, 63 , 2 2, , [email protected], Phone number-8105418762 / 7019881906, , 13

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 10. If S is the sum of the first 10 terms of the series, , (a) 0, , 1, 1,  1 , tan    tan 1    tan 1    ...,, 3, 7,  13 , Then tan  S  is equal to:, 1, , 6, (a), 5, , 5, (b), 6, , 10, (c), 11, ,  x 2  xy  xy  y 2 , , , y  x  y,   tan 1 1  ,  tan 1  2, 2,  y  xy  x  xy , 4, , , y  x  y, , , , 1, 1, 1,  tan 1  tan 1  ...10th term, 3, 7, 13,  2 1 ,  3 2 ,  S  tan 1 ,  tan 1 , ,  1  2 1 ,  1  3  2  , , , , ,  43 ,  11  10 ,  tan 1 ,  ...  tan 1 , ,  1  4  3 ,  1  1110  , , , , , 1, 1, 1, 1,  S   tan 2  tan 1   tan 3  tan 2 , S  tan 1, , , , , , , , TYPE:2 SIMPLIFIED FORM, LEVEL-I, , 1., , sin 1, , x, is equal to, xa, , x, a, x, a, , (a) cos 1, (c) tan 1, , (b) cosec1, , x, a, , (d)None of these, , Solution:(c) Putting x  a tan 2 , , sin 1, , x, a tan 2 , a tan ,  sin 1,  sin 1, 2, xa, a sec , a tan   a, ,  x,  sin 1 sin     tan 1 ,  ., a, , , , x, 1  x  y , 2. The simplified form of tan    tan , ,  y,  x y, 1, ,  3  5cos x ,  is equal to,  5  3cos x , x, x, 1, , (a) tan 1  tan , (b) 2 tan 1  2 tan , 2, 2, 2, , 1, x, x, , 1, (c) tan 1  2 tan , (d) 2 tan 1  tan , 2, 2, 2, , 2, , 3. cos 1 , ,  11  1 ,  S  tan 1 11  tan 1 1  S  tan 1 , ,  1  11 ,  10 , 5,  S  tan 1    tan 1  ,  12 , 6, ,  5   5 ,  tan  S   tan  tan 1      ,  6   6 , , , , , (d) , ,  x x y , , , , x,  x y, y x y , 1 , tan 1    tan 1 , , tan, ,  x  x  y ,  y,  x y, 1  , , y  x  y  , , , Solution:(b), , , , 2, , Solution:(b), , [JEE Main-05/09/2020-Shift-I-4M], [One option correct type], , , , (c) , , 4, , [KCET-2016-1M] [One option correct type], [NCERT-DIRECT QUESTION], , 5, (d), 11, ,  tan 1 4  tan 1 3  ...  tan 1 11  tan 1 10, , (b) , , Solution:(d)We take x , , , then cos x  0, 2, ,  3  5cos x , 1  4 , 1  3 , cos 1 ,   cos    tan  3 ,  ,  5  3cos x , 5, , x, 1, Put x  in 2 tan 1  tan  we get, 2, 2, 2, , 1, 2 tan 1  tan , 4, 2, ,  1 ,  2. 2 , 4, 1  1 , 1,  tan 1   .,  2 tan    tan , , 1, 3, 2,  1 , 4, , LEVEL-II, ,  1  x2  1  x2 , 1, , x  ,, 2, 2, 2,  1  x  1  x , , 1. The values of tan 1 , , x  0 is equal to,  1 1 2, (a)  cos x, 4 2, , 1 2, (c)  cos x, 4, , (b), (d), ,  1 1 2,  cos x, 4 2, , 1 2, 4, ,  cos x, , [JEE-Main-4m- online-17], , Solution:(b), , is equal to, 14, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, ,  1 x  1 x , 2, tan 1 ,  Put x  cos 2, 2, 2,  1  x  1  x , cos 1 x 2, x 2  cos 2  cos 1 x 2  2 , , 2, 2, , ,  tan 1 , , ,  tan 1 , , , , 3. Considering only the principal values, if, , 2, , 1  cos 2  1  cos 2, 1  cos 2  1  cos 2, , , , , , 2cos 2   2sin 2  ,  cos   sin  ,   tan 1 , , 2, 2,  cos   sin  , 2cos   2sin  , ,  1  tan  , , , 1 ,  tan 1 ,   tan  tan     ,  1  tan  , ,  4,  1, , 1, 2,      cos x, 4 2, 4, ,  , , TYPE:3 SOLVE THE INVERSE TRIGONOMETRIC, EQUATIONS, LEVEL-I, 1, 1, 1. If tan 2 x  tan 3 x , , (a) – 1, (c) 1,, , , , 4, , , then x =, (b), , 1, 6, , 1, 6, , (d) None of these, , Solution:(b) Trick : Check with the options., , 1, Obviously the equation holds for x  ,, 6, but not for – 1., , ,  1 , tan(cos 1 x)  sin cot 1    , then x is equal to,  2 , , 5, 1, 2, 3, (a), (b), (c), (d), 3, 5, 5, 5, 1, 1, Solution:(d)Put cot 1      cot  , 2, 2, 2,  sin  , . Put cos 1 x   then x  cos , 5, 2, 5, ,  x  cos  , ., 3, 5, 11, 4. The equation 2 cos 1 x  sin 1 x , has, Also  tan  , , 6, (a)No solution, (b)Only one solution, (c)Two solutions, (d)Three solutions, Solution:(a) Given equation is, 11, 2 cos 1 x  sin 1 x , 6, 11, 1, 1,  cos x  (cos x  sin 1 x) , 6, 11, , , 4,  cos 1 x , , which is not, ,  c o s1 x , 6, 2, 3, possible as cos 1 x 0,   ., , 5., , 2 tan 1 (cos x)  tan 1 (cosec2 x), then x =, (a), , , 2, , (b) , , (c), , , 6, , (d), , , 3, , 1 , 2. If sin x  cot   , then x is, 2 2, 3, 1, 2, a)0, (b), (c), (d), 2, 5, 5, 1 , Solution:(b) sin 1 x  cot 1   , 2 2, , Solution:(d) 2 tan 1 (cos x)  tan 1 (cos ec 2 x), 2 cos x , 1 , 1 ,  tan 1 ,   tan  2 , 2, 1, , cos, x, sin, x, , , , , , 1 1, 1 1 ,  cot 2  cos, , 5, , , 6. If tan 1 x  tan 1 y  tan 1 z   , then, 1 1 1,   , xy yz zx, , 1, , 1, , 1 ,  ;, 5 2, , sin 1 x  cos 1, , 1, Clearly, x , 5, , 2, , , 1, , CHETHAN M G, , , , 2 cos x, 1, , 2, sin x sin 2 x, ,  2 cos x  1  x , , , 3, , ., , 1, (d) xyz, xyz, Solution:(b) If tan 1 x  tan 1 y  tan 1 z  , then x  y  z  xyz, (a)0, , (b)1, , (c), , x  y  z  xyz Dividing by xyz, we get, 1, 1, 1,  , 1., yz xz xy, [email protected], Phone number-8105418762 / 7019881906, , 15

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CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , 2, 1 1,  sin 1  sin 1 x, then x is equal to, 7. If sin, 3, 3, 5 4 2, (a) 0, (b), 9, , 54 2, (c), (d), 9, 2, , 10. If 3 tan 1 x  cot 1 x   then x is, , x  y 1, 2, , 2, , , , 8. sin  sin 1, , , a) 1, , [KCET-16-1m] [Only one option correct], , c), , 4, 5, , d), , 3 ,  , 4, 4, 4 , , 4, ,  , , (a) 1  y, , (c) 1, , 1, (d), 3, , [KCET-15-1m] [Only one option correct], [NCERT-DIRECT QUESTION], ,  1 x  1, 1,   tan x ,  x  0 ,  1 x  2, 1,  tan 11  tan 1 x   tan 1 x, 2,  1,   tan 1 x  tan 1 x , 4 2,  3, 2,   tan 1 x ,  tan 1 x, 4 2, 12, 1, , ,   tan 1 x  x  tan, x, 6, 6, 3, , Solution:(d) tan 1, , 16, , (b) y, , 2, , 2, , (c) 0, , 1, , Solution:(a) sin x  sin, , (d) 1  y, , 1, , y, , , ,, 2, , ,  sin 1 y, 2, 1,  sin x  cos 1 y  sin 1 y  sin 1 y   , ,  sin 1 x , , , , , 2 , ,  sin 1 x  sin 1 1  y 2  cos1 y  sin 1 1  y 2 , , , , , (b) 1, , 2, , , then x 2 is equal to, , [KCET-16-1m] [Only one option correct], , Solution:(d) sin  sin 1, , (a) 3, , , , 11. If sin 1 x  sin 1 y , , 1, 5, , 1, ,  cos 1 x   1 ;, 5, , 1, , 1, sin–1  cos 1 x   x , 5, 2, 5,  1 x  1, 1, 9. Solve for x tan 1 ,   tan x, x  0 is,  1 x  2, , 1, 2, , 3tan 1 x  cot 1 x    1, Put option (b) x  1 in (1), Eq 1  3tan 1 1  cot 1 1  , , 1, ,  cos 1 x  = 1 then x =, 5, , , b) 0, , (d), , Solution:(b) By inspection method, , Solution:(c), , 1, 1, 2, 4 2, 1, sin 1  sin 1  sin 1  1   1  , 3, 3, 9 3, 9, 3,  54 2,  sin 1 , , 9, , , 54 2, Therefore, x , ., 9,  sin 1 x  sin 1 y  sin 1{x 1  y 2  y 1  x 2 }, , , 2, 2, If 1  x, y  1 and x  y  1 or if xy  0 and, , (c) 1, , (b) 1, , (a) 0, , , , , ,  x  1  y Squaring on both sides, 2, ,  x2  1  y 2, 12., , The number of real solutions of, , tan 1 x( x  1)  sin 1 x 2  x  1 , (a)Zero, (c)Two, , , 2, , is, , (b)One, (d)Infinite, , Solution:(c) tan 1 x( x  1)  sin 1 x 2  x  1 , , tan, , 1, , , 2, , x( x  1) is defined, when x( x  1)  0, , sin 1 x 2  x  1 is defined, when, 0  x( x  1)  1  1 or 0  x( x  1)  0 ........(ii), From (i) and (ii), x ( x  1)  0 or x  0 and –1., Hence, number of solutions is 2.,  x 2 x3, , ,  , x4 x6, 13. If sin 1  x    ...   cos 1  x 2    ...  , 2 4, 2 4, , , ,  2, for 0  | x | , (a) 1, 2, , 2, then x equals, 1, (b)1, (c) , 2, , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , (d) – 1, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , Solution:(b), , We know that, , sin 1 y  cos 1 y , 2, , , , 2, , Solution:(b), , , | y | 1  According to question,, , , , 3, , x, x2, ,  2 x  x3  2 x 2  x3  x  x 2, 2  x 2  x2, ,  x  x 2  0  x(1  x)  0  x  0 and x = 1,, but x  0. So, x  1 ., 2, 2x, 1 2a, 1 1  b, , cos,  tan 1, , then, 14. If sin, 2, 2, 1 a, 1 b, 1  x2, x, , a b, 1  ab, Solution:(d) Put a  tan  , b  tan  and x  tan ,, (a) a, , (b) b, , (c), , ab, 1  ab, , (d), , then reduced form is, sin 1 (sin 2 )  cos 1 (cos 2 )  tan 1 (tan 2 ),  2  2  2      , , tan   tan ,  tan, 1  tan  .tan , , a b, x, 1  ab, , 2, , 2, , cos1 x  cos1 y  cos1 z  cos1 t  , , , , cos1 ( x)  0,  , , , , x  y  z  t  cos( )  1,  x  y 2  z 2  t 2  (1)2  (1)2  (1)2  (1) 2  4, 2, 2x, 2x, , 1 1  x, , 4cos,  2 tan 1, , 16. If 3sin 1, 2, 2, 2, 1 x, 1 x, 1 x, 3, then x , 2, , (a) 3, (c)1, CHETHAN M G, , 1, 3, (d) None of these, (b), , 3(2 )  4(2 )  2(2 ) , , , , 3, , , ,  6  8  4 , , , , , , 3, , 3, , 1,    tan 1 x   x  tan , 6, 6, 6, 3, , 17. For the equation cos x  cos 2 x    0 ,, the number of real solution is, (a) 1, (b) 2, (c) 0, (d) , 1, 1, Solution:(c) cos x  cos (2 x)  , 1, , 1, ,  cos 2 x    cos x,  2 x  cos(  cos 1 x), 1, , 1, ,  2x   x  x  0, But x  0 does not satisfy the given equation., , 1. If sin, , the value of x  y  z  t, a) xy  zy  zt, b) 1 2 xyzt, c) 4, d) 6, 1, 1, Solution:(c) cos x  cos y  cos 1 z  cos1 t  4, This is only possible if, 2, , , , , , LEVEL-II, , cos1 x  cos1 y  cos1 z  cos1 t  4 , then, 2, , 3sin 1 (sin 2 )  4cos 1 (cos 2 ) 2 tan 1 (tan 2 ) , , No solution will exist., , Substituting these values, we get, 15. If, , 2,  2 tan  , 1  1  tan  , 3sin 1 , , 4cos, , 2 , 2 ,  1  tan  ,  1  tan  ,  2 tan   , 2 tan 1 , , 2,  1  tan   3, ,  2 x  cos  (cos cos 1 x)  sin  sin(cos 1 x), , Taking tan on both sides, we get tan(   )  tan, , , 2, 2x, 2x, , 1 1  x, , 4cos,  2 tan 1, , 2, 2, 2, 1 x, 1 x, 1 x, 3, , Putting x  tan , , x, x, x 4 x6,   ...  x 2 , ,  ..., 2, 4, 2, 4, x, x2, , , ( 0  | x | 2 ), , 2, x, x, 1, 1, 2, 2, x, , 3sin 1, , 1, ,  x 2   sin, , on, (a) circle, (c) ellipse, , 1, , y3   2, , then p(x, y) lies, , (b) parabola, (d) hyperbola, ,  a   sin, , 1, Solution:(c) If sin x, , 1, , yb  , , then, , x 2 2 xy, y 2  sin 2 , , cos, , , b2, a 2 ab, a  2, b  3,   , 2, 2, 2, x, 2 xy,   y  sin 2 , , cos, 2 9, 2, 22 (2)(3), x2 y 2, ,  1  p ( x, y ) lies on ellipse, 4 9, 1  12 , 1  16 , 2. If x  0, cos      cos   then ' x ', 2,  x,  x, equals, (a) 12 16, , (b) 4, , 3, , (c) 3, , 4, , (d) 20, , [email protected], Phone number-8105418762 / 7019881906, , 17

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Solution:(d) If cos, , 1, , A  cos 1 B , , CONCEPTS OF MATHEMATICS FOR KCET -JEE MAIN AND ADVANCED, , , , 2, , then, , 2, , A  B 1, , 1  12 , 1  16 , So cos    cos   ,  x,  x 2, 2, , 2, , 2, , 2, , 2, 2,  12   16 ,        1  12  16  x2,  x  x,  x  20  x  0 , , 3. A value of x satisfying the equation, sin cot 1 1  x   cos  tan 1 x  is, , 1, 2, , (a), , (c) 1, , (b) 0, , (d) , , 5. Considering only the principal values of inverse, functions, the set, , 1, 2, , , , A   x  0 : tan 1 2 x  tan 1 3 x  , 4, , , [JEE-Main-4m- online-17], , Solution:(d) cot, , 1, , 1  x     1  x  cot , , tan 1  x     x  tan , 2  2x  x2, , 1, , , , 1 x, , sin cot, , 1, , a) Contains two elements, b) Contains more than two elements, c) is an empty set, d) is a singleton, , 1  x2, , x, , [JEE-MAIN-12/1/2019-SHIFT-I], [One option correct type-4M], , , 1, , 1  x   cos tan, , , , 1, sin sin 1 , 2, ,  2  2x  x, 1, 1, , 2, 2  2x  x, 1  x2, , 1, , Solution:(d) tan, , x , ,  1  1, ,    cos cos , 2, ,  ,  1 x, , , ,  , , 1, 2, 3,  2 ,  3  , 4. If cos 1    cos 1 ,   , x  then x , 4,  3x ,  4x  2, 2  2x  x2  1  x2  x  , , a) 145, , b) 145, , 11, , 12, , c), , 146, 10, , d), , 146, 11, , [JEE-Main-4m- online-19], ,  2 ,  3 ,  cos 1    cos 1 1   ,  3x ,  4x , , 1, , 2 x  tan 1 3 x , , , 4, ,  2 x  3x  ,  tan 1 , ,  1   2 x  3x   4, , , ,  5x ,  5x  , ,  tan,  tan 1, , 2 , 2 , 4,  1 6x ,  1 6x  4, 5x, ,  1  5x  1  6 x2, 1  6 x2, 2,  6x2  5x 1  0  6 x  6 x  x  1  0,  6 x  x  1  1 x  1  0,   x  1 6 x  1  0  x  1 or x , , 1, 6, , 1,  x  0, 6, A is a singleton, 6. All x satisfying the inequality, x, ,  2 ,  3  , Solution:(b) cos    cos 1 , ,  3x ,  4x  2,  2  ,  3 ,  cos 1     cos 1  ,  3x  2,  4x ,  2 ,  3 ,  cos 1    sin 1  ,  3x ,  4x , 1, , 18, , 2, 2, 9,  3 , ,  1   ,  1, 3x, 3x, 16 x 2,  4x , 4, 9, 4, 9,  2  1,  2, 1, 2, 9x, 16 x, 9 x 16 x 2, 64  81, , 1, 9  16  x 2, 145, 145,  x2 , x, 9  16, 3 4, 145 , 3, x, x , , 12 , 4, ,  cot x , 1, , 2, ,  7 cot 1 x  10  0, lie in the interval., , a)  ,cot 5    cot 4,cot 2 , , 2, ,  cot 5, cot 4 , c)  , cot 5   cot 2,  , d)  cot 2,  , b), , [email protected], https://t.me/mcpmathematicskar_1986, Phone number-8105418762 / 701988190, , MCP-MATHEMATICS

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INVERSE TRIGONOMETRIC FUNCTIONS, , [JEE-MAIN-11/1/2019-SHIFT-II], [One option correct type-4M], , , , Solution:(d) cot 1 x, , , , , , , , 2, ,  7 cot 1 x  10  0,, , , ,  cot 1 x  5 cot 1 x  2  0, , 10, ,  cot 1 x   , 2    5,  , , 5,  cot 1 x   0, 2   cot 1 x   0,  ,  x   cot 2,  , , 2, , 5, ,  y,    , where, 2, y, 1  x  1,  2  y  2, x  , then for all, 2, 2, 2, x, y, 4 x  4 xy cos   y is equal to:, , 7. If cos 1 x  cos 1 , , 2, a) 4sin , 2, b) 2sin , c) 4sin 2   2x 2 y 2, , 2, 2 2, d) 4 cos   2x y, , [JEE-MAIN-10/4/2019-SHIFT-II], [One option correct type-4M], ,  y,   ,, 2, , Solution:(a) cos 1 x  cos 1 , , , ,  cos 1   cos 1   cos 1   1   2 1   2, , , if  1  x, y  1 & x  y, , ,  , , , ,  xy, y2 ,  cos 1 ,  1  x2 1 ,  ,  2, , 4, , , 1  x2 4  y 2, xy, , ,  cos , 2, 2, ,  xy  1  x2 4  y 2  2cos ,  1  x2 4  y 2  2cos   xy, Squaring on both sides, we get, , , , , , , ,  1  x2 4  y 2  4cos2   x2 y 2  4 xy cos ,  4  y  4 x  x 2 y 2  4cos 2   x 2 y 2  4 xy cos , 2, , 2, , , ,  4 x2  4 xy cos   y 2  4 1  cos2 , , , ,  4 x  4 xy cos   y  4sin , 2, , CHETHAN M G, , 2, , 2, , [email protected], Phone number-8105418762 / 7019881906, , 19