Title 10. INVERSE TRIGANOMETRIC FUNCTIONS MEDIUM ANS.pdf ✅

1. (b) 3 3 sin sin 1    = −            − −       −   − 2 sin 2  1   x 2. (c) o o o sec [sec( 30 )] sec (sec 30 ) 30 1 1 − = = − − . 3. (c) 2 3 3 sin 3 5 sin sin 1  1  = −          = −     −  − 4. (d) The principal vlaue of )] 3 2 sin [sin( 1   − − = 3 3 sin sin 1    =     −  5. (d) Put 2 1 cot 2 1 cot 1  =  =     −     . 5 2 sin = Put =  − x 1 cos then x = cos  Also  3 5 , cos 5 2 tan = x =  = 6. (a) sin [sin( 600 )] 1 o = − −  = sin [ sin(360 240 )] 1 − + −  sin [ sin 240 ] 1 o = − −  sin [ sin(180 60 )] 1 o o = − + −  o sin sin60 −1  = 3 3 sin sin 1    =            = − 7. (a)        +     −  − 3 5 sin sin 3 5 cos cos 1  1               + −            = − − − 3 sin sin 2 3 cos cos 2 1 1     0 3 3 = − =   8. (a) Given equation is 6 11 2 cos sin 1 1  + = − − x x  6 11 cos (cos sin ) 1 1 1  + + = − − − x x x  6 2 11 cos 1   = − − x 3 4 cos 1   = − x , which is not possible as   − cos 0, 1 x . 9. (b) 3 2 sin sin 1 1  + = − − x y  3 2 cos 2 cos 2  1  1  − + − = − − x y  3 cos cos 1 1  + = − − x y 10. (b) We know that , | | 1 2 sin cos 1 1 + =  − − y y y   According to question, .......... ... 2 4 .......... 2 4 4 6 2 2 3 − + − = − + − x x x x x x  , ( 0 | | 2 2 1 2 1 2 2   + = + x x x x x  )  2 2 2 2 x x x x + = +  3 2 3 2x + x = 2x + x  2 x = x 0 (1 ) 0 0 2 x − x =  x − x =  x = and x = 1, but x  0. So, x =1 11. (b) 2 2 sin cot 1 1   =      + − − x x         = − − 5 1 cos 2 1 cot 1 1  5 2 1 sin cos 1 1  + = − − x ; Clearly, 5 1 x = 12. (d) sin(cot ) 1 x − =         + − 2 1 1 1 sin sin x = 2 1 1 + x 13. (c) 2 tan ( 1) sin 1 1 1 2  + + + + = − − x x x x tan ( 1) 1 + − x x is defined, when x(x + 1)  0 ........(i) sin 1 1 2 + + − x x 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 14. (d)                  +     −  − 13 3 cos 5 3 tan sin 1 1 =       + − − 3 2 tan 4 3 tan tan 1 1 =             − + − 3 2 . 4 3 1 3 2 4 3 tan tan 1 =        − 6 12 12 17 tan tan 1 = 6 17 15. (b) 4 tan 1 3 1 . 2 1 1 3 1 2 1 tan 3 1 tan 2 1 tan 1 1 1 1  = = − + + = − − − − 16. (c)         + =         + = − + − − − − − 9 5 4 2 sin 9 1 1 3 2 9 4 1 3 1 sin 3 2 sin 3 1 sin 1 1 1 1 Therefore, 9 5 + 4 2 x = 17. (b) 1 2 
cot 3 5 1 5 1 1 cot 3 cot 5 1 sin−1 −1 −1 −1 +               − + = = 4 cot (1) 3 2 2 3 1 cot (2) cot (3) cot 1 1 1 1   = =      +  − + = − − − − 18. (d) Given, 13 12 cos 5 3 sin sin −1 −1 −1 C = +  13 5 sin 5 3 sin sin −1 −1 −1 C = +         = − + − − 25 9 1 13 5 169 25 1 5 3 sin 1 65 56 65 56 sin 1   =      = − C 19. (c) 16 63 tan 4 3 tan 5 12 tan −1 −1 −1 + + = 16 63 tan 20 36 48 15 tan −1 −1 + − +  + ( xy  1 ) =  − + =  − − 16 63 tan 16 63 tan 1 1 20. (a)         = − + − − 25 16 1 3 1 9 1 1 5 4 sin 1  =         + =         + − − 15 8 2 3 sin 15 3 15 8 2 sin 1 1 Since 2 1, 15 8 2 3     + 2 ( ) 3 1 sin 5 2 4 sin 2 1 1        = −       = − + − − −     21. (a) + = − − 3 cos 2 cos 1 x 1 y  cos 9 1 4 1 3 . 2 2 2 =         −         − − x y x y  ( 6 cos ) (4 )(9 ) 2 2 2 xy −  = − x − y     2 2 2 2 9x − 12 xy cos + 4y = 36(1 − cos ) = 36 sin 22. (b) x x 1 1 1 sin 2 3 sin 2 sin − − − = −         = − − − − 4 3 . 1 1 2 3 sin 1 2 x x  2 1 2 3 2 2 x x = − x − (1 ) 4 3 2 5 2 2 x x  = −       or 28 3 2 x =  7 3 2 1 28 3 x = = , (not 7 3 2 1 − ) 23. (d) Since, 2 1 1 1 2 2 tan tan x x x − = − −  25 1 1 5 2 2 tan 5 1 2 2 tan 5 1 4 tan 1 1 1 − =      = − − − 576 100 1 24 20 tan 24 10 2 tan 1 1 − = = − − 119 120 tan−1 = So, 239 1 tan 119 120 tan 239 1 tan 5 1 4 tan−1 −1 −1 −1 − = − 239 1 . 119 120 1 239 1 119 120 tan 1 + − = − (119 239 ) 120 (120 239 ) 119 tan 1  +  − = − 4 tan 1 1  = = − 24. (d) If x = −1, LHS = , 2  RHS =        − 2 2  . So, the fomula does not hold. If x  −1, the angle on the LHS is in the second quadrant while the angle on the RHS is 2 (angle in the fourth quadrant), which cannot be equal. If x  1, the angle on the LHS is in the second quadrant while the angle on the RHS is 2 (angle in the first quadrant) and these two may be equal. If −1  x  0 , the angle on the LHS is positive and that on the RHS is negative and the two cannot be equal 25. (a) Let x = tan . Then sin (sin 2 ) 1 tan 2 tan sin 1 2 sin 1 2 1 2 1   − −  − = + = + x x  2 sin (sin 2 ) 1 2 2 tan sin 1 2 1 1   − − − = + + + x x x If 2 1 1 1 2 , 2 tan sin 2 2 2 x x x + −   +  − −   = + =  − x 1 2 2 4 tan independent of x. If 2 1 1 1 2 , 2 tan sin 2 2 2 x x x + −  −  +  − −    = 2 sin [sin( 2)] 2  2 1 + − = + − − =  = independent of x.         − 4 , 4    but        4 3 , 4    and from the principal value of x 1 tan − .        − 2 , 2    . Hence,        2 , 4            2 , 4     =  + + − − 2 1 1 1 2 tan sin x x x . Also at       = + = + = + = + − − − 2 2 sin 1 4 2. 1 2 , 2 tan sin 4 1 2 1 1 x x x .  The given function =  = constant if        2 , 4    . i.e., x [1,+ ) 26. (d) If we denote x 1 cos− by y, then Since    − 0 cos x 1  0  2y  2 .........(1) Also since 2 sin 2x 1 x 2 1 2          −  − −
 ( ) 2 sin sin 2y 2 1     − −  2 2y 2     − ...........(2) From (1) and (2) we find 2 0 2y     4 0 y     4 0 cos x 1    − which holds if x 1 2 1   27. (a) Given equation can be written as tan–1 (x – 1) + tan–1 (x + 1) = tan–1 3x – tan–1 x  tan–1 1 (x 1)(x 1) x 1 x 1 − − + − + + = tan–1 2 1 3x 3x x + −  2 2 1 3x 2x 2 x 2x + = −  x + 3x3 = 2x – x 3  4x3 – x = 0  x(4x2 – 1) = 0  x = 0, x = ± 1⁄2 none of which satisfies 1 < x < 2 28. (b) (cot–1 x – 1) (cot–1 x – 2) > 0  x < cot 2 or x > cot 1 (cot–1 x is a decreasing function) 29. (d) tan cos–1 5 4 = 4 3 , tan tan–1 3 2 = 3 2  expression in the question = 1 3/ 4 2 / 3 3/ 4 2 / 3 −  + = 6 17  Choice (d) is correct. 30. (b) By tan (A + B + C) formula tan (tan–1  + tan–1 + tan–1 ) = tan −    +  +  −    − − − − − − − − 1 1 1 1 1 1 1 1 1 tan tan tan tan tan (tan )(tan )tan = tan 1 3 ( m) ( m) − − − − = tan 0 = 0  tan–1  + tan–1 + tan–1  = n 31. (a) tan2 (sin–1 x) > 1  4  < sin–1 x < 2  or – 2  < sin–1 x < – 4  x          ,1 2 1 or x          − − 2 1 1,  x          − − 2 1 1,          , 1 2 1 Hence (A) is the correct answer. 32. (b) cos (2cos–1 x + sin–1 x) = cos = –sin (cos–1 x) = – sin(sin–1 ) = – 2 1− x = – 2 6 1 1       − = – 36 35 = – 6 35 . 33. (d)  > 2tan–1x  0 2  > tan–1x  0 also 1 + x2  1 – x 2  x  0 34. (b) x x x x 1 1 1 1 tan 2 sin cos tan − − − − = + − = −   We know 2 tan 2  1  −   − x 4 tan 2 0 2 tan 2  1   1    −  −   −  − − x x 35. (a) We have =                  −         − − − 2 2 2 2 1 cos . 1 1 b y a x b y a x 1 1 cos 2 2 2 2 =         −          − − b y a x ab xy  2 2 2 2 2 2 2 2 2 cos 1 a b x y b y a x ab xy  = − − +      −  2 2 2 2 2 2 2 2 2 2 2 2 2 cos 1 2 cos a b x y b y a x ab xy a b x y +  −  = − − +     2 2 2 2 2 2 cos 1 cos sin 2 − + = − = b y ab xy a x 36. (a) a b c a b c ca b a b c b c a a b c ( ) tan ( ) tan ( ) tan 1 1 1 + + + + + + + + = − − −  Let abc a b c s + + = 2  1 2 2 1 2 2 1 2 2 tan a s tan b s tan c s − − −  = + +  tan ( ) tan ( ) tan ( ) 1 1 1 as bs cs − − −  = + +          − − − + + − = − 2 2 2 3 1 1 tan abs bcs cas as bs cs abcs   0 1 ( ) ( ) tan 2 2 =         − + + + + − = ab bc ca s a b c abcs  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 3 tan 3 tan 3 , 1 1 1  = + + = − − − tan = 0 37. (d)       +  − x 1 cos 2 2 1− x
2 tan (cos ) 1 x − tan (cos ec ) 1 2 x − =  x x x x x x 2 2 2 1 2 1 sin 1 sin 2cos sin 1 tan 1 cos 2cos tan   =       =      − − −  3 2 cos 1  x =  x = 38. (c) x x 1 1 sin 2 tan − − =  2 1 1 1 2 sin sin x x x + = − −  x x x = + 2 1 2  0 3 x − x =  x(x + 1)(x −1) = 0  x = {−1,1, 0} 39. (b)        +      − − − x x 1 1 2 tan 2 tan 2 sin  = 2 sin  = 1 40. (d) Put a = tan,b = tan and x = tan, then reduced form is sin (sin 2 ) cos (cos 2 ) tan (tan 2 ) 1 1 1    − − − − =  2 − 2 = 2   −  = Taking tan on both sides, we get tan( − ) = tan       tan 1 tan .tan tan tan = + − Substituting these values, we get x ab a b = + − 1 41. (d)             − −  =       −     −  − − tan (1) 25 1 1 5 2 tan tan 5 4 1 tan 2 tan 1  1 1 = 17 7 12 5 1 1 12 5 tan (1) tan.tan 12 5 tan tan 1 1 1 − =             + − =       − − − − 42. (c) 99 1 tan 70 1 tan 5 1 4 tan−1 −1 −1 − + = 99 1 tan 70 1 tan 25 1 1 5 2 2 tan −1 −1 −1 − +             − = 99 1 tan 70 1 tan 12 5 2 tan−1 −1 −1  − +      = 99 1 tan 70 1 tan 144 25 1 6 5 tan −1 −1 −1 − +             − = 99 1 tan 70 1 tan 119 120 tan−1 −1 −1  − +      =             + − + − − 70 1 . 99 1 1 70 1 99 1 tan 119 120 tan 1 1 =       + − − − 6931 29 tan 119 120 tan 1 1 = 239 1 tan 119 120 tan 6931 29 tan 119 120 tan−1 −1 −1 −1 − = − = 4 tan (1) 239 1 119 120 1 239 1 119 120 tan 1 1  = =             +  − − − 43. (b) cos[tan (2 2)] 3 1 sin 2 tan−1 −1  +            = cos[tan (2 2)] 9 1 1 3 2 sin tan −1 −1 +             − cos[tan 2 2] 4 3 sin tan−1 −1 +      = = 15 14 3 1 5 3 3 1 cos cos 5 3 sin sin 1 1 = + =       +      − − 44. (b) Let b a b a =  = − cos  cos 1       + −       + − − b a b 1 a 1 cos 2 1 4 cos tan 2 1 4 tan   = t t t t + − + − + 1 1 1 1 , where 2 tan  t = a b t t 2 cos 2 1 (1 ) 2 2 2 = = − + =  45. (b) tan 3 1 tan tan −1 −1 −1 + = y x or y 1 tan −1 = x 1 1 tan 3 tan − − − or x x y 1 3 3 tan 1 tan 1 1 + − = − −  x x y − + = 3 1 3 As x, y are positive integers, x =1 , 2 and corresponding y = 2, 7  Solutions are (x, y) = (1, 2),(2,7) 46. (d) sin x cos x tan x −1 −1 −1  = + −  tan x 2 −1 −   = and since 4 0 tan x 1    − when 0  x  1 we fined 4 2      47. (d) The given expression is equal to ( )       +  + = − − − sin x 2 cos cos x sin x cos 1 1 1 = ( ) 5 1 sin sin x x 1 − = − = − − 48. (c) The given expression is equal to        − −  6 tan 3 2 tan cos ec 1 = 6 3 2 x 3 1 2 tan 3 1 3 2 2 tan 1 1  =  = =         − − −