Title 10. INVERS TRIGANOMETRIC FUNCTIONS HARD ANS.pdf ✅

1. (c) p + − p = − − q − − − cos 1 4 3 cos cos 1 1 1  1  p p − − q         − + − = − − − − cos 1 2 1 cos cos 1 cos 1 1 1 1          p − p − − p p = − − q − . q 2 1 . 1 2 1 1 1  0 = 1 − q − q  2 1 q = 2. (c) Let = =  − − a b b 1 a 1 tan , tan   a b b a tan = , tan  =        +      − − a b b b a c a 2 1 3 2 1 3 tan 2 1 sec 2 tan 2 1 osec 2 =       2 2 sin2 3  a +       2 2 cos2 3  b = 2 2 3 2 2 3 3 3 1 1 1 cos 1 cos a b a b a b b a b a + + + + − = + + −   =         + − + − + + − + + + 2 2 2 3 2 2 2 2 2 3 2 2 2 2 ( ) [ ] ( ) [ ] a b a b a b a a b b a a b b a b (rationalized) [ { } { }] 2 2 2 2 2 2 = a + b a a + b + b + b a + b − a = [ ( )] ( )( ) 2 2 2 2 2 2 a + b a + b a + b = a + b a + b 3. (d) We have          + −       −      − − − − 2 1 3cos 2 1 sin 2 1 cos 1 1 1 - 4 tan ( 1) 1 − − = 12 25 4 4 4 3 3 x 6 2 x 3 2   =       − −  +  −  4. (b) 4 3 cot 2 x 7 7 1 cot 5 3 2cot 7 cos 1 2 −1 −1 −1 − + −  =      + = 4 3 cot 7 24 cot−1 −1 + = 4 3 7 24 1 4 3 x 7 24 cot 1 + − − 117 44 cot−1 = 117 125 cosec−1 = 5. (c) A = 2tan–1 ( 2 2 – 1) > 2 tan–1 3 = 1200 B = 3tan–1 2 2 1 + cos–1 5 4 < 3tan–1 ( 2 −1) + cos–1 4 5 +1 = 2 1 103 6. (c)                − −  − 6 7 sin sin 2 1 =        +  − 6 7 sin sin 2 1 = 2 6  −  = 3  7. (c) For         −   − , 2 3 , tan  < 0  tan–1 (cot ) – cot–1 (tan ) = –  Also for points in 2nd quadrant sin–1 (sin ) + cos–1 (cos ) = . 8. (d) xy > 0  x & y are of same sign x + x 1  2 or  – 2  y 2/3 /2 /3 x –2 –1 1 2 sec–1       + x 1 x         2 , 3          3 2 , 2  z          , 3 2          3 4 , Hence, value of z (among the given) which does not lie in the set is 3 5 . 9. (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 10. (c) For x          ,2 2 3 , cos–1 (cos x) = cos–1 (cos (2 – (2 – x))) = cos–1 (cos (2 – x)) = 2 – x and sin–1 (sin x) = sin–1 (sin (2 + (x – 2))) = x – 2  cos–1 (cos x) + sin–1 (sin x)
= (2 – x) + (x – 2) = 0. Therefore sin–1 (cos(cos–1 (cos x) + sin–1 (sin x))) = sin–1 0 = 0 11. (c) Since x y x y . y x − + > 1 .The given expression is equal to  + tan–1             − + −  − + + x y x y y x 1 x y x y y x =  + tan–1 (x y ) x y 2 2 2 2 − + + =  + tan–1 (–1) = 3/4. 12. (c) Let tan–1 1/3 =  and tan–1 2 2 = . Then tan  =1/3 and tan  = 2 2 , so that sin (2 tan–1 (1/3)) + cos (tan–1 2 2 ) = sin 2 + cos  = +   2 1 tan 2 tan + +  2 1 tan 1 [ –/2 <  < /2] = 1 (1/ 9) 2.(1/ 3) + + 1 8 1 + = 15 14 3 1 5 3 3 1 10 9 . 3 2 + = + = . 13. (a) Take tan on both sides x x 1 1 x 1 x 1 + − − + = – 7  x 2 – 4x + 4 = 0  x = 2 check L.H.S = tan–1 3 + tan–12 (0, ) R.H.S. (0, ) 14. (d) [cot–1 x] = 0, 1, 2, 3, [tan–1 x] = –2, –1, 0, 1 Case (i) [cot–1 x] = [tan–1 x] = 1 x  (cot 2, cot 1], & x  [tan 1, )  x  as cot 1 < tan 1 (ii) [cot–1 x] = 3, [tan–1 x] = – 1 x  (–, cot 3], x  [– tan1, 0]  x  as cot 3 < – tan 1 (iii) [cot–1 x] = 2, [tan–1 x] = 0 x (cot 3, cot 2], x [0, tan 1) x as cot 2 < 0 So no solution 15. (c) f(x) = 2  + 2  + sec–1 x But 0  sec–1 x  , sec–1 x  2     f(x)  2, f(x)  2 3 . 16. (a) tan–1 2 1 6x 2x 3x − + = 4   5x = 1 – 6x2  6x2 + 5x – 1 = 0  x = –1, 1/6 But sum of the –ve number cannot be /4, hence x = 1/6 is only solution. 17. (c) Since 5 2 2 3    , We have sin 5 < 0, so sin–1 (sin 5) = 2 – 5 Thus the given inequality can be written as 2 – 5 > x2 – 4x or x2 – 4x – (2 – 5) < 0          − −  − − 2 4 16 4(2 5) x         + −  − − 2 4 16 4(2 5) x < 0  [x – 2 – 9 − 2 ] [x – (2 + 9 − 2 )] < 0 x  (2 – 9 − 2 ), (2 + 9 − 2 ). 18. (a)  – 1  x – 3  1  2  x  4 ..........(i) Also – 1  x – 1  1 0  x  2 .........(ii) form (i) & (ii) x = 2 Also, 2 – x   0 x  ± 2  At x = 2 sin–1 (2–3) + cos–1 (2–1) + tan–1       2 − 4 2 = cos–1k –  2  − + 0 4  − = cos–1 k –   cos–1 k = /4  k = 2 1 19. (a) cosec (cosec–1x) = x  x  R –(–1, 1) also range of cosec–1 (cosec x)         − ,0 2         2 0, so combining these two. x        −  − , 1 2         2 1, 20. (b) Using tan–1x + tan–1y = tan–1         − + 1 xy x y 21. (a) sin  = cos (sin–1 x) = cos       −  − cos x 2 1 = sin (cos–1 x)  cos  = sin (cos–1 x)
 sin  = cos   tan  = cot  22. (c) cos–1 x = tan–1 x = cos–1 2 1 x 1 +  x = 2 1 x 1 +  x 2 = 2 −1+ 5  2 x 2 = 4 5 −1 cos–1         2 x 2 = 5 2 23. (c) since x y x y y x − +  > 1 given sum =  + tan–1             − + − − + + x y x y . y x 1 x y x y y x =  + tan–1 (–1) = 3/4 24. (c) We have A = 2 tan–1 (2 2 – 1) = 2 tan–1 (1.828)  A > 2 tan– 1 3  A > 3 2 Next sin– 1       3 1 < sin–1       2 1  sin– 1       3 1 < 6   sin– 1 3 1 < 2  Also 3 sin– 1       3 1 = sin–1               − 3 3 1 4 3 1 3. = sin–1       27 23 = sin–1 (0. 852)  3 sin–1       3 1 < sin–1         2 3  3 sin–1       3 1 < 3  Further sin–1       5 3 = sin–1 (0. 6) < sin–1         2 3  sin–1       5 3 < 3  Hence, B = 3 sin–1       3 1 + sin–1       5 3 < 3  + 3  = 3 2 . Hence A > B 25. (b) L.H.S. of choice (B) is a negative number and R.H.S. is a positive number. 26. (b) Given equation is 3.2 tan–1 x – 4.2 tan–1 x + 2.2 tan–1 x = 3   2 tan– 1 x : 3   x = tan 6  = 3 1 27. (c) – 2   sin–1 x  2   sin–1 x + sin–1 y + sin– 1 z = 2 3  sin–1 x = sin–1 y = sin– 1 z = 2   x = y = z = 1 Also f(p + q) = f(p). f(q)  p, q  R ...(1) Given f(1) = 1 from (1), F(1 + 1) = f(1). F(1)  f(2) =12 = 1 from (2), f(2 + 1) = f(2) . f(1)  f(3) = 12 . 1 = 13 = 1 Now given expression = 3 – 3 3 = 2 28. (b) We have x1 = sin 2, x1 x2 = cos 2, x1 x2 x3 = cos  and x1 x2 x3 x4 = – sin   tan–1 x1 + tan–1 x2 + tan–1 x3 + tan–1 x4 = tan–1 1 2 1 2 3 4 1 1 2 3 1 x x x x x x x x x x −  +  −  = tan–1 −  −   −  1 cos2 sin sin 2 cos = tan–1  −   −  (2sin 1)sin (2sin 1)cos = tan–1 (cot ) = tan–1 tan (/2 – ) = /2 – . 29. (c) Let cos–1       b 3a =  cos  =       b 3a Now tan        +  4 2 + tan        −  4 2 = 2 1 tan 2 1 tan  −  + + 2 1 tan 2 1– tan  +  =                       + 2 1– tan 2 2 1 tan 2 2 = cos 2 =       b 3a 2 =       a b 3 2 . 30. (b) Adding both equation we get (sin–1 y)2 =       + 32 4n 1  2  0  32 4n +1  2  4 2 
 – 4 1  n  8 7 Also cos–1 x = 32 4n −1  2  0     − 2 32 4n 1  4 1  n   8 + 1. 31. (d) cos–1 (cos 4) = cos–1 {cos (2 – 4)} = 2 – 4  cos–1 (cos 4) > 3x 2 – 4x  2 – 4 > 3x 2 – 4x  3x 2 – 4x – (2 – 4) < 0  3 2 6 8 x 3 2 6 8 + −   − − . 32. (a) Let tan = tan x, then u = cot–1 (tan x) – tan–1 (tan x) = 2  – x – x = 2  – 2x  2x = 2  – u  x = 4  – 2 u  tan x = tan       −  2 u 4  tan = tan       −  2 u 4 . 33. (c) tan–1 m m 2 2m 4 2 + + = tan–1 1 (m m 1)(m m 1) 2m 2 2 + + + − +  tan–1 (m2 + m + 1) – tan–1 (m2 – m + 1) So that  = n m 1 –1 tan         m + m + 2 2m 4 2 = (tan–1 3 – tan–1 1) + (tan–1 7 – tan–1 3) + .... + tan–1 (n2 + n + 1) – tan–1 (n2 – n + 1)  tan–1 (n2 + n + 1) – tan–1 (1)  tan–1         + + + n n 2 n n 2 2 = (tan–1 3 – tan–1 ) + (tan–1 7 – tan–1 3) + .....+ (tan–1 (n2 + n + 1) – tan–1 (n2 – n + 1)) = tan–1 (n2 + n + 1) – tan–1 (n2 – n + 1) = tan–1 n n 2 n n 2 2 + + + . 34. (c) Graphs of y = sin–1 (sin x) and |y| = cos x meet exactly six times in [–3, 3]. Y O x –2 2 –3 2 5 − 2  − 2 3 2 3 − 2  2 5 3 35. (d) – 2   sin–1 x  2   sin–1 xi = 2   xi = 1, 1 < i < 1000  = 1000 i 1 i x = 1000. 36. (c) Let cos–1 p = , cos–1q = , cos–1 r =  cos  = p, cos  = q, cos  = r , +  +  =  cos ( + ) = cos ( – )  cos .cos  – sin  sin  = –cos   pq – 2 2 1− p 1− q = – r  p 2 + q2 + r2 + 2pqr = 1  p 2 + q2 + r2 + 2pqr + 4 = 5. 37. (c) Clearly, x(x + 1)  0 and x 2 + x + 1  1. Together they imply x (x + 1) = 0.  x = 0, –1. When x = 0, L.H.S. = tan–1 0 + sin–1 1 = 2  . When x = –1, L.H.S. = tan–1 0 + sin–1 1−1+1 = 0 + sin–1 1 = 2  , Thus two solution. 38. (d) cos  = 1/x, tan  =   cos sin = 1/ x 1 (1/ x) 2 − = x 1 2 − 39. (a) Let cos–1 x = . Then x = cos   tan  = sec 1 2 − = 1/ x 1 2 − = 2 1− x /x 40. (b)  tan–1 x + tan–1 y =  – tan–1 z  x +y/1–xy = – z  x + y + z = xyz divide by xyz, we get 1/xy + 1/yz + 1/zx = 1 41. (a) Let cos–1 4/5 = x  cos x = 4/5 ...(i) Now sin ( 5 4 cos 2 1 −1 ) = sin (x/2) ...(ii)