Chapter 29 Inverse Trigonometrical Functions PDF

60 126 Inverse Trigonometrical Functions The inverse of a function f : A B exists if f is one-one onto i.e., a bijection and is given by f (x )  y  f 1 (y)  x. E3 Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a bijection and so it is not invertible. If...

60 126 Inverse Trigonometrical Functions The inverse of a function f : A B exists if f is one-one onto i.e., a bijection and is given by f (x )  y  f 1 (y)  x. E3 Consider the sine function with domain R and range [–1, 1]. Clearly this function is not a bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes one–one, then it would become invertible. If we consider sine as a function with domain     2 , 2  and co-domain [–1, 1], then it is a bijection and therefore, invertible. The inverse of   ID    sine function is defined as sin 1 x    sin   x , where    ,  and x  [1, 1].  2 2 5.1 Properties of Inverse Trigonometric Functions. (1) Meaning of inverse function cot   x  cot 1 x   (iv) U (ii) cos   x  cos 1 x   (i) sin   x  sin1 x   (iii) tan   x  tan 1 x   (v) sec   x  sec 1 x   (vi) cosec  x  cosec 1 x   D YG (2) Domain and range of inverse functions 1 (i) If sin y  x , then y  sin Y (1, /2) x , under certain condition.  1  sin y  1; but sin y  x.  1  x  1 Again, sin y  1  y    2 and sin y  1  y   2. Keeping in mind numerically smallest angles or real numbers.   O  y = sin– 1 x X  U  y(–1, – 2 /2) 2 These restrictions on the values of x and y provide us with the domain and range for the function y  sin 1 x. ST i.e., Domain : x  [1, 1] Range:    y   ,   2 2 Let cos y  x , then y  cos (ii)  Y 1 x , under certain conditions  1  (–1, cos y 1 /2) 1  x  1 cos y  1  y   cos y  1  y  0 O y = cos– 1 X x (1, 0)  0  y   {as cos x is a decreasing function in [ 0 ,  ]; hence cos   cos y  cos 0 These restrictions on the values of x and y provide us the domain and range for the function y  cos 1 x. Inverse Trigonometrical Functions 127 i.e. Domain: x  [1, 1] y  [0,  ] Range : If tan y  x , then y  tan 1 x , under certain conditions. Here, tan y  R  x  R ,    tan y      2 y y  y= 2 2 60 (iii) Thus, Domain x  R ;    Range y    ,   2 2 y = –/2 y = tan– 1 x E3 (iv) If cot y  x , then y  cot 1 x under certain conditions, cot y  R  x  R; y= ID    cot y    0  y   These conditions on x and y make the function, cot y  x one-one and onto so that the inverse function exists. i.e., y  cot 1 x (0, /2) is O y = cot– 1  Domain : x  R y  (0,  ) D YG (v) If sec y  x , then y  sec 1 x , where | x |  1 and 0  y   , y  x  (–1,) 2 Here, Domain: x  R  (1, 1) Range:   y  [0,  ]    2 If cosec y  x , then y  cosec 1 x U (vi) Where | x |  1 and   2 y  2 ST Here, Domain  R  (1, 1)    Range   ,   {0}  2 2 Function sin 1 x x U meaningful. Range : x O y = /2 y O (1, /2) ,y  0 (1,0 – y= ) sec 1 x x (–1, – ) Domain (D) 1  x  1 or [1, 1] cos 1 x 1  x  1 or [1, 1] tan 1 x   x   i.e., x  R or (, ) cot 1 x   x   i.e., x  R or O y = cosec– 1 x Range (R)   2    2    ,   2 2 or  0     or [0,  ]       or   ,  2 2  2 2 0     or (0,  )   x 128 Inverse Trigonometrical Functions (, ) x  1, x  1 or (,  1]  [1, ) x cosec 1 x  x  1, x  1 or (,  1]  [1, ) (3) sin 1 (sin  )   , Provided that   2    2      0,   2   sec 1 (sec )   , Provided that 0    cosec 1 (cosec )   , Provided that  (4) sin(sin 1  2 2  2      , 0    0,   2   2 or  Provided  2      0 or 0     2 cos(cos 1 x )  x, Provided that  1  x  1 x )  x, Provided that  1  x  1 , tan (tan 1 x )  x , Provided that    x   x  cot(cot 1 x )  x, that U Provided sec(sec 1 x )  x, Provided that    x  1 or 1  x   D YG cosec (cosec –1 x )  x, Provided that    x  1 or 1  x   (5) sin 1 ( x )   sin 1 x cot 1 ( x )    cot 1 x (6) sin 1 x  cos 1 x   2 sec 1 x  cosec -1 x  cos 1 ( x )    cos 1 x , tan 1 ( x )   tan 1 x sec 1 ( x )    sec 1 x cosec 1 ( x )  cosec –1 x tan 1 x  cot 1 x  , for all x  [1, 1]  2  2 , for all x  R , for all x  (,  1]  [1, )  Here; sin Here; cos 1 1 x , cosec x , tan x , sec 1 x , cot 1 1 /2 x belong to I and IV Quadrant. x belong to I and II Quadrant. ST  U Important Tips 1   I Quadrant is common to all the inverse functions. III Quadrant is not used in inverse function.  IV Quadrant is used in the clockwise direction i.e.,  II I   2 IV y0 – (7) Principal values for inverse circular functions Principal values for x  0 0  sin 1 x  0  cos 1 x   2 that cot 1 (cot  )   , Provided that 0     , or  E3 2    2 ID   cos 1 (cos )   , , 0   tan 1 (tan  )   , Provided that   , 0     or 0,    ,   2  2 2  60 sec 1 Principal values for x  0    2 2  2  sin 1 x  0  cos 1 x   I 0 Inverse Trigonometrical Functions 129 0  cot 1 x  0  sec 1 x  1  2  2  2 4  2 2   2  2  tan 1 x  0  cot 1 x    sec 1 x    cosec 1 x  0 4 2 3  5 1   1  2 Thus sin 1    , not not ; tan 1 ( 3 )   not ; cot 1 (1)  not ; cos 1     6 3 3 4 3 2 6  2 3 E3  2  0  cosec x     60 0  tan 1 x  etc. Note :  sin 1 x , cos 1 x , tan 1 x are also written as arc sin x , arc cos x and arc tan x respectively. ID  It should be noted that if not otherwise stated only principal values of inverse circular functions are to be considered. x 1 x2  cot 1  1 x2 1  sec 1   2 x  1x D YG sin 1 x  cos 1 1  x 2  tan 1 U 1 1 (8) Conversion property : Let, sin 1 x  y  x  sin y  cosec y     y  cosec –1   x x    cosec –1  1   x   1 x2 cos 1 x  sin 1 1  x 2  tan 1   x    1   sec 1 1  cosec –1    2 x  1x   x tan 1 x  sin 1   2  1 x 2     cot 1  1   sec 1 1  x 2  cosec 1  1  x   x x   1  sin 1    cosec 1 x , for all x  (,1]  [1, ) x U Note :     cos 1  1   2   1 x   x   cot 1    2   1x      1 1 cos 1    sec 1 x , for all x  (, 1]  [1, ) x x ST      1 for x  0  1   cot x , tan 1     1  x     cot x , for x  0 – x2 (9) General values of inverse circular functions: We know that if is the smallest angle whose sine is x, then all the angles whose sine is x can be written as nx  (1)n , where n  0, 1, 2,..... Therefore, the general value of sin 1 x can be taken as n  (1)n . The general value of sin 1 x is denoted by sin 1 x. Thus, we have sin 1 x  nπ  (1) n α,  1  x  1, if sin α  x and  π π α 2 2 Similarly, general values of other inverse circular functions are given as follows: 130 Inverse Trigonometrical Functions cos 1 x  2n   ,  1  x  1 ; If cos   x , 0     tan 1 x  n  , x  R ; If tan   x,  cot 1 x  n    , x  R ; If cot   x , 0     sec 1 x  2n    , x  1 or x  1 ; If sec   x , 0     and  cosec1 x  n  (1)n, x  1 or x  1 ; If cosec  x ,  Solution: (b)      sin 1 sin     3 3    Example: 2 sec 1[sec(30 o )]  (b)  30 o 5   The principal value of sin 1  sin  is 3   D YG Example: 3 (b)  5 3 Solution: (c)  5  3    sin 1  sin .   sin 1     3  2 3    Example: 4   2 The principal value of sin 1 sin   3 2 3 U (a)  Example: 5 1 (b) 5 60    2 2 and x  0 4 3 (c) 30 (d) 2 (c)  o  3 5 8 [MP PET 1992] (d) 150 (d) 2 3 [IIT 1986] (c) 4 3 (d) None of these 2    )] = sin 1 sin  . 3 3 3 2 5 (c) 3 5. Put cos 1 x   then x  cos  2 5 1 ,  x  cos   o 4 3   is  5 Also  tan   [Roorkee 1992] [MP PET 1996] 1 1 Put cot 1      cot   2 2  sin   Example: 6    1  Considering only the principal values, if tan(cos 1 x )  sin cot 1   , then x is equal to  2   (a) Solution: (d) (b) The principal vlaue of sin 1 [sin(  ST Solution: (d) (c) o sec 1 [sec( 30 o )]  sec 1 (sec 30 o )  30 o. 5 3 2      sin 1 x   2 2  Solution: (c) (a) 2 U (a) – 60   E3  3   is The principal value of sin 1   2    2  (a)  (b)  3 3   2 ID Example: 1  5. 3 If   sin [sin ( 600 )] , then one of the possible value of  is (d) 5 3 Inverse Trigonometrical Functions 131 (a) Solution: (a)  3  2 (b) (c) 2 3 (d)  2 3   sin 1 [sin(600 o )] = sin1[ sin(360  240 )]         sin 1 [ sin 240 o ]  sin 1 [ sin(180 o  60 o )]    sin1 sin 60 o  sin 1 sin  .   3  3 5  5    Value of cos 1  cos   sin 1  sin  is 3  3    (a) 0 [Roorkee 2000]  2 (b) (c) 60 Example: 7 2 3 (d)     5  5          cos 1  cos   sin 1  sin   cos 1 cos  2    sin 1 sin 2      0. 3 3 3 3 3 3           Example: 8 The equation 2 cos 1 x  sin 1 x  Given equation is 2 cos 1 x  sin 1 x   cos 1 x  If sin 1 x  sin 1 y  (a) Solution: (b)  Example: 10  2  cos 1 x   2    6 (d)   2  cos 1 y  2 3  cos 1 x  cos 1 y   3. 3 4 (b) 0    U ST  2  tan 1 x   2  (c)  4  4   0 (d) 1 2    tan 1 x   2  2 0   2  tan 1 x  We know that sin 1 y  cos 1 y   x 1 x 2  (c)  (b) 1  According to question, x   4    2  tan 1 x 2    4.      x4 x6 x2 x3 ...   cos  1  x 2   ...   for 0 | x |  2 , then x equals If sin  1  x   2 4 2 4     2 (a) Solution: (b) (c) 2 3   sin 1 x  cos 1 x  tan 1 x  We know  Example: 11  3 [EAMCET 1994] If   sin 1 x  cos 1 x  tan 1 x , x  0 then the smallest interval in which  lies is (a) Solution: (b) 2 , then cos 1 x  cos 1 y  3 (b) sin 1 x  sin 1 y  (d) Three solutions 4 , which is not possible as cos 1 x  0,  . 3  cos 1 x  2 3 [AMU 1999] 11 11  cos 1 x  (cos 1 x  sin 1 x )  6 6 D YG Example: 9 11   6 2 (c) Two solutions U Solution: (a) (b) Only one solution ID (a) No solution 11 has 6 E3 Solution: (a) 10  3  2 1 2 [IIT Screening 2001] (d) – 1 , | y| 1 x2 x3 x4 x6  ..........  x 2   ............. 2 4 2 4 x2 x x2  , ( 0  | x |  2 )   2x  x 3  2x 2  x 3  x  x 2 2 2  x 2  x2 x 1 2 132 Inverse Trigonometrical Functions  x  x 2  0  x (1  x )  0  x  0 and x = 1, but x  0. So, x  1. Example: 12 If sin 1 x  cot 1  1    , then x is 2 2 [Karnataka 1999; Roorkee 1999] 1   5 2 5 2 (d) 5  1 1    cot 1  cos 1 2 5   x  sin 1 x  cot 1    2 2 sin 1 x  cos 1 (c) ; Clearly, x  1 5 3 2  2. E3 Solution: (b) 1 (b) 60 (a) 0 1 Example: 13 The value of sin(cot 1 x ) is [MP PET 2001; UPSEAT 1987] (b) (1  x 2 )3 / 2 Example: 14  sin(cot 1 x ) = sin sin 1   1x 2   =   1 1 x2 The number of real solutions of tan 1 (a) Zero tan 1 x (x  1)  sin 1. x (x  1)  sin 1 x2  x 1  (b) One x2  x  1   2 is (c) Two [IIT Screening1999] (d) Infinite  D YG Solution: (c) 1 U Solution: (d) (d) (1  x 2 )1 / 2 (c) (1  x 2 )1 / 2 ID (a) (1  x 2 )3 / 2 2 tan 1 x(x  1) is defined, when x ( x  1)  0........(i) 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. 5.2 Formulae for Sum and Difference of Inverse Trigonometric Function. If x  0, y  0 and xy  1  xy   ; (2) tan 1 x  tan 1 y    tan 1   1  xy  If x  0, y  0 and xy  1  xy   ; (3) tan 1 x  tan 1 y    tan 1   1  xy  If x  0, y  0 and xy  1  x y   ; (4) tan 1 x  tan 1 y  tan 1   1  xy  If xy  1  x y   ; (5) tan 1  tan 1 y    tan 1   1  xy  If x  0, y  0 and xy  1  x y   ; (6) tan 1 x  tan 1 y    tan 1   1  xy  If x  0, y  0 and xy  1 ST U  x y   ; (1) tan 1 x  tan 1 y  tan 1   1  xy  Inverse Trigonometrical Functions 133  x  y  z  xyz  (7) tan 1 x  tan 1 y  tan 1 z  tan 1    1  xy  yz  zx   S  S 3  S 5 ...........  (8) tan 1 x 1  tan 1 x 2 ..........  tan 1 x n  tan 1  1  , 1  S 2  S 4  S 6 ........  (10) xy  1 yx cot 1 x  cot 1 y  cot 1 xy  1 yx E3 (9) cot 1 x  cot 1 y  cot 1 60 where S k denotes the sum of the products of x 1 , x 2 ,........, x n taken k at a time. (11) sin 1 x  sin 1 y  sin 1 {x 1  y 2  y 1  x 2 } ; (12) ID If  1  x , y  1 and x 2  y 2  1 or if xy  0 and x 2  y 2  1 sin 1 x  sin 1 y    sin 1 {x 1  y 2  y 1  x 2 }, If 0  x , y  1 and x 2  y 2  1 (14) U (13) sin 1 x  sin 1 y    sin 1 {x 1  y 2  y 1  x 2 }, If  1  x ; y  0 and x 2  y 2  1 sin 1 x  sin 1 y  sin 1 {x 1  y 2  y 1  x 2 }, If D YG and x 2  y 2  1. (15) sin 1 x  sin 1 y    sin 1 {x 1  y 2  y 1  x 2 }, (16) If  1  x , y  1 and x  y  0. cos 1 x  cos 1 y  2  cos 1 {xy  1  x 2 1  y 2 } , If  1  x , y  1 and x  y  0 cos 1 x  cos 1 y  cos 1 {xy  1  x 2 1  y 2 }, If  1  x , y  1, and x  y. cos 1 x  cos 1 y   cos 1 {xy  1  x 2 1  y 2 }, If  1  y  0, 0  x  1 and x  y. ST (20) xy  0 If 0  x  1,  1  y  0 and x 2  y 2  1. U (19) or sin 1 x  sin 1 y    sin 1 {x 1  y 2  y 1  x 2 }, If  1  x  0, 0  y  1 and x 2  y 2  1. (17) cos 1 x  cos 1 y  cos 1 {xy  1  x 2. 1  y 2 } , (18)  1  x ; y  1 and x 2  y 2  1 if Important Tips   If tan 1 x  tan 1 y  tan 1 z   If tan 1 x  tan 1 y  tan 1 z   , then x  y  z  xyz.  If sin 1 x  sin 1 y  sin 1 z   If sin 1 x  sin 1 y  sin 1 z   , then x 1  x 2  y 1  y 2  z 1  z 2  2 xyz.  If cos 1 x  cos 1 y  cos 1 z  3 , then xy  yz  zx  3. 2  2 , then xy  yz  zx  1. , then x 2  y 2  z 2  2 xyz  1. 134 Inverse Trigonometrical Functions  If cos 1 x  cos 1 y  cos 1 z   , then x 2  y 2  z 2  2 xyz  1.  If sin 1 x  sin 1 y  sin 1 z   If sin 1 x  sin 1 y   , then cos 1 x  cos 1 y    .  If cos 1 x  cos 1 y   , then sin 1 x  sin 1 y    .  If tan 1 x  tan 1 y   If cot 1 x  cot 1 y   If cos 1  2 60 2 , then xy  1. , then xy  1. x 2 2 xy y2 x y cos   2  sin 2 .  cos 1   , then 2  ab a b a b   3  3  is The value of tan sin 1    cos 1  5   13  6 17 (a) 13 13 5 (c) (d) [AMU 2001] 17 6   3  3 2  3  = tan  tan 1  tan 1  tan sin 1    cos 1  4 3  5   13  U Solution: (d) 6 (b) ID Example: 15 E3  3 , then xy  yz  zx  3. 2 Example: 16 D YG 3 2      17 12  17    = = tan  tan 1 4 3  = tan  tan 1. 3 2  12 6 6   1.   4 3  tan 1 2001] 1 1  tan 1  2 3 (a) 0  4 (c)  2 (d)  1 1  1 1 1 1 2 3  tan 1 1  . tan  tan  tan 1 1 4 2 3 1. 2 3 1 2 If sin 1  sin 1  sin 1 x , then x is equal to 3 3 1 U Solution: (b) (b) [MP PET 1997, 2003; UPSEAT 2003; Karnataka CET ST Example: 17 (a) 0 Solution: (c) sin 1 5 4 2 9 (c) (d)  2 1  5 4 2 1 2 4 2 1  sin 1  sin 1  1  1    sin 1   3 3 3 9 3 9 9     Therefore, x  Example: 18 5 4 2 9 (b) [Roorkee 1995] 1 sin 1 5 5 4 2. 9  cot 1 3 is equal to [MP PET 1993; Karnataka CET 1995] (a)  6 (b)  4 (c)  3 (d)  2 Inverse Trigonometrical Functions 135   1 1  5  cot 1 3  cot 1  1 5   5  1 Example: 19 If sin 1 65 56 (a)  sin 1 C  sin 1 Example: 20 24 65 (b) Given, sin 1 C  sin 1 (c) 2  2 (b) f    2 cos 1  3 3 3   56 65 = cos 1 x  (cos 1      1  (c) f    3 3 1  1 (d) f    2 cos 1  3 3 3   U  3 1. 1 x2 Soltuion: (a,d) f (x )  cos 1 x  cos 1  x  2 2   1 1  cos 1 x ) , according as cos 1  or  cos 1 x 2 2 D YG 1 2 1 if cos 1  cos 1 x , which holds for x  2 3 2 = 2 cos 1 x  cos 1 sin 1 (d)  25 5 9  56 3 5 3  1  56   sin 1  1  1  sin 1.   sin    C  65 5 13 169 13 25    65  5  x 1  3  3 x 2 , then If f (x )  cos 1 x  cos 1   2 2  = cos 1 16 65 3 12  cos 1 5 13 2  (a) f    3 3 Example: 21 [Pb. CET 1999] ID Solution: (d) 3  12   cos 1    sin 1 C, then C  5  13  60 sin 1 E3 Solution: (b)     1 1 1  2  3  1  1 1   cot 3 = cot (2)  cot (3)  cot  3  2   cot (1)  4.      1 1 1 if cos 1  cos 1 x , which holds for x . 2 2 3 12 4 63  cos 1  tan 1  13 5 16 (a) 0 (b)  2 (c)  (d) 3 2 12 48  15 63 3 63 63 63  tan 1  tan 1  tan 1  tan 1 . =   tan 1 ( xy  1 ) =   tan 1 5 20  36 16 4 16 16 16 tan 1 Example: 22 If   sin 1 ST U Solution: (c) 4 1 4 1  sin 1 and   cos 1  cos 1 , then 5 3 5 3 (a)    Solution: (a) 4   sin 1   5 Since (b)    1 (c)    (d) None of these    3  1 1 16  1 8 2 1 8 2  3    1   sin   = sin   9 3 25  15   15  15  8 2 3   1,   15 2 4  1    sin 1   sin 1   (   )     . 5 2 3 2 2   Example: 23 If cos 1 x y  cos 1   , then 9 x 2  12 xy cos   4 y 2  2 3 (a) 36 sin 2  (b) 36 cos 2  (c) 36 tan 2  (d) None of these 136 Inverse Trigonometrical Functions Solution: (a) cos 1  x y  cos 1   2 3  x y x 2   y 2 .   1  1   cos  2 3 4   9   The number of solutions of sin 1 x  sin 1 2 x  (a) 0 (b) 1 sin 1 2 x  sin 1  2x  3 2 1  x2  x 2 2 (d) Infinitie  x 3 1  28 2 3 1 , (not  7 2 3 ). 7 x 2 2 xy y2 x y If cos 1    cos 1     , then 2  cos   2  ab a b a b [UPSEAT 1999; MP U PET 1995] (a) sin 2  (c) tan 2  (b) cos 2     x y x2   y2  We have cos 1 .  1  2  1  2      a b a   b      D YG Solution: (a) (c) 2  3 3 3. 1  x2  x 1    sin 1 x  sin 1  2 4   2 3  5x  2    (1  x 2 ) or 28 x  3 2 4   Example: 25 is 3 E3 Solution: (b)  ID Example: 24 60  (xy  6 cos  ) 2  (4  x 2 )(9  y 2 )  9 x 2  12 xy cos   4 y 2  36 (1  cos 2  )  36 sin 2 .  (d) cot 2   xy x2   y2    1  2   1  2   cos  ab a   b   2 x 2 y 2 x 2y 2   xy  cos    1  2  2  2 2  a b ab   ab  x 2y 2 2 a b  cos 2   2 xy x2 y2 x 2y 2 cos   1  2  2  2 2 ab a b a b x2 a 2  2 xy y2 cos   2  1  cos 2   sin 2 . ab b (a) 0   tan 1 Let s 2  a(a  b  c)  tan 1 bc b(a  b  c) ca c(a  b  c) , then tan  is ab ST  tan 1 Solution: (a)  If a, b, c be positive real numbers and the value of   tan 1 U Example: 26 2 a(a  b  c)  tan 1 bc abc abc (c) a  b  c (b) 1 b(a  b  c)  tan 1 ca (d) None of these c(a  b  c) ab    tan 1 a2 s 2  tan 1 b 2 s 2  tan 1 c 2 s 2    tan 1 (as)  tan 1 (bs)  tan 1 (cs)   (a  b  c)  abcs 2 tan   s  2  1  (ab  bc  ca)s  0   as  bs  cs  abcs 3    tan 1  2 2 2  1  abs  bcs  cas    [ abcs 2  (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 Inverse Trigonometrical Functions 137   tan 1 3  tan 1 3  tan 1 3   , All possible values of p and q for which cos 1 cos 1 Solution: (c)  1 2 (b) q  1, p  p  cos 1 1  p  1 2 (c) 0  p  1, q  1 2 3 holds, is 4 (d) None of these  1    cos 1 1  q p  cos 1 1  p  cos 1    2 3  cos 1 1  q  cos 1 4 60 (a) p  1, q  p  cos 1 1  p  cos 1 1  q   1  1 1 p 1  p  1  p p  . 1q . q  0  1q  q  q . 2  2  2 5.3 Inverse Trigonometric Ratios of Multiple Angles. 1 1 x 2 2 (2) 2 sin 1 x    sin 1 (2 x 1  x 2 ), If 1 ID (1) 2 sin 1 x  sin 1 (2 x 1  x 2 ) , If  E3 Example: 27 tan   0. (3) 2 sin 1 x    sin 1 (2 x 1  x 2 ) , If  1  x  (4) 3 sin 1 x  sin 1 (3 x  4 x 3 ), If 2 1 x   1  x 1 2 1 2 D YG (7) 2 cos 1 x  cos 1 (2 x 2  1) , If 0  x  1 (9) 3 cos 1 x  cos 1 (4 x 3  3 x ) If  1 1 x 2 2 1  x 1 2 (11) 3 cos 1 x  2  cos 1 (4 x 3  3 x ) , If  1  x   ST U  2x  (13) 2 tan 1 x    tan 1   , If x  1 2 1  x  x  1  2x (15) 2 tan 1 x  sin 1  2 1  x   , If  1  x  1  1 2 (8) 2 cos 1 x  2  cos 1 (2 x 2  1) , 3 cos 1 x  2  cos 1 (4 x 3  3 x ) , (12)  2x  2 tan 1 x  tan 1   , if  1  x  1 1  x 2  (14)  2x  2 tan 1 x    tan 1  , 2 1  x   2x (16) 2 tan 1 x    sin 1  2 1  x 1  x 2 (18) 2 tan 1 x  cos 1  2 1  x (19) 1  x 2 2 tan 1 x   cos 1  2 1  x  1 1  , If x  3 3  if  1  x  0 (10)  2x  (17) 2 tan 1 x    sin 1   , If x  1 2 1  x   3x  x 3 3 tan 1 x  tan 1  2  1  3x 1 1 x 2 2 (6) 3 sin 1 x    sin 1 (3 x  4 x 3 ), If U (5) 3 sin 1 x    sin 1 (3 x  4 x 3 ) , If 1  x 1 2 If If   , If x  1    , If 0  x       , If    x  0   (20) 138 Inverse Trigonometrical Functions (21)  1  , If x   3   x tan 1  2  a  x 2 (24)  1  , If x    3  (23)  1 x   sin a  E3  3a2 x  x 3  x tan 1  2  3 tan 1 2  a  a(a  3 x )   1  x2  1  x2 tan 1   1  x 2  1  x 2 1x 1  cos 1 x 1 x 2 2 tan 1 (cos x )  tan 1 (cosec 2 x ), then x = (a) Solution: (d)  2 (b) 2 tan 1 (cos x )  tan 1 (cos ec 2 x ) D YG  1  cos x   6 (b) {1, 2} Soltuion: (c) sin 1 x  2 tan 1 x  sin 1 x  sin 1 Exmaple: 30  1  x2  sin tan 1    2x  U 2     cos 1  1  x  1  x2   ST (a) 0 Solution: (b)    sin   2 tan 1 x   2 tan 1 2    Example: 31 If sin 1 (a) a (26) [UPSEAT 2002] (d)  3 The solution set of the equation sin 1 x  2 tan 1 x is (a) {1, 2} Solution: (d) (c)   1    cos 1 x 2  4 2  1  2 cos x 1  2 cos x  1  x .   2 3 sin 2 x sin 2 x  sin x   tan 1  2 cos 2x   tan 1  Example: 29  U Example: 28 ID (25) tan 1 (22) 60  3x  x 3 3 tan 1 x    tan 1  2  1  3x  3x  x 3 3 tan 1 x    tan 1  2  1  3x 2x 1 x2  2x 1 x2 [AMU 2002] (c) {1, 1, 0} 1 (d) {1, , 0} 2  x  x 3  x  0  x(x  1)(x  1)  0  x  {1, 1, 0}.   is equal to    (b) 1 [Kurukshetra CEE 2001] (c) 2 (d) 1 2   x  = sin = 1. 2  2 2a 2x 1 1  b  cos  tan 1 , then x  2 2 1a 1b 1  x2 (b) b [EAMCET 1989] (c) ab 1  ab Put a  tan  , b  tan  and x  tan  , then reduced form is sin 1 (sin 2 )  cos 1 (cos 2 )  tan 1 (tan 2 )  2  2  2       Taking tan on both sides, we get tan(   ) = tan   tan   tan   tan  1  tan . tan  (d) ab 1  ab Inverse Trigonometrical Functions 139 Substituting these values, we get Example: 32 ab x 1  ab  1   tan  2 tan 1      5 4  (a) 17 7 [IIT 1984] (b)  17 7 (c) 7 17 (d)  7 17 Example: 33 4 tan 1 1 1 1  tan 1  tan 1  5 70 99  2 (b) 4 tan E3 Solution: (c) 1  3 (c)  2  1 1 1 1 1 1  tan  tan = 2 tan  5 5 70 99 1  1  25  4 U   1 1 1  tan 1   tan 70 99   1  1  99  70  1 1 1 1 1 120 1  tan  tan  = tan  = tan   tan 70 99 119  119  1  1. 1  99 70 D YG 1  120 = tan Example: 34   29   1 120  tan 1     = tan 119  6931    1  120  119  239 120 1 1 29 1 120 1 1  tan  tan  tan = tan  119 6931 119 239  1  120  1  119 239    1   tan (1) . 4     1  The value of sin  2 tan 1     cos (tan 1 2 2 )   3   16 15 U (a) (b) 14 15 [AMU 1999] (c) 12 15 (d) 11 15 2    1 3   1  1  1 1 sin 2 tan    cos [tan (2 2 )] = sin  tan   cos [tan (2 2 )] 1 3     1   9  ST Solution: (b) (d) None of these   1 1 1  tan 1   tan 70 99    5  5  1 1 1 1 1  tan = 2 tan    tan = tan  6 70 99  12   1  25  144 1  1 [Roorkee 1981] ID (a) 60 Solution: (d) 2    5  1     1 5   5   1  1  1   7. tan  2 tan      tan  tan  tan (1) = tan  tan 1  tan 1 (1)  tan. tan 1  12 1 5   17 5 4 12         1 1     25 12   3 3 1  3 1 14     sin  tan 1   cos[tan 1 2 2 ] = sin sin 1   cos cos 1    . 5 3  5 3 15 4      Example: 35 a a  1  1 tan   cos 1   tan   cos 1  equal to 4 2 b 4 2 b     (a) Solution: (b) 2a b Let cos 1 (b) 2b a [MP PET 1999] (c) a b (d) b a a a    cos   b b a a 1  t 1  t   1  1 tan   cos 1   tan   cos 1  =  , where t  tan b b 1  t 1  t 2 4 2 4 2 2 (1  t 2 ) 2 2b  . cos  a 1  t2 140 Inverse Trigonometrical Functions  (a) Solution: (d) 1 1  tan 1 is equal to 5 239 (b) Since, 2 tan 1 x  tan 1  4 tan So, 4 tan Example: 37  3 (d)  4 2x 1 x2 1 1 120 1  tan 1  tan 1  tan 1 5 239 119 239 The formula cos 1 1 x2 1 x2 (a) x  R  tan 1 120  119 120 1 119 1 239  tan 1 (120  239 )  119 1 (119  239 )  120. 239  2 tan 1 x holds only for (c) x  (1, 1] (b) | x |  1  tan 1 1   4. (d) x  [1,  ) ID Solution: (d) (c) 2 20 120 1  1 1  1 5 1 10 1 24  tan 1  2 2 tan  2 tan  tan   2 tan 100 1 119 5 5 24   1 1 576 25 1 1  2 60 4 tan 1 E3 Example: 36    , RHS = 2    . So, the fomula does not hold. 2  2 If x  1, the angle on the LHS is in the second quadrant while the angle on the RHS is 2  (angle in If x  1, LHS = U 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 Solution: (a) D YG Example: 38 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. 2x is independent of x , then 2 tan 1 x  sin 1 1 x2 (a) x  [1,  ) (b) x [1, 1] (c) x  (,  1 ] (d) None of these Let x  tan . Then sin 1  2 tan 1 x  sin 1   2x 1 x 2 2x 1 x 2  sin 1 2 tan  1  tan 2   sin 1 (sin 2 )  2  sin 1 (sin 2 ) 2x = 2  2  4 tan 1 x  independent of x. 1  x2   2x If     2  , 2 tan 1 x  sin 1 = 2  sin 1 [sin(  2 )]  2    2 =  = independent of x. 2 2 1  x2  2  , 2 tan 1 x  sin 1 U If  2 2 ST  3       ,  but    ,  and from the principal value of tan 1 x.  4 4 4 4        , . Hence,    ,   2 2 4 2         ,   tan 1 x  sin 1 4 2 Also at    4 , 2 tan 1 x  sin 1 2x 1 x2 2x 1 x2  2. .  4  sin 1 1   2   2 .     The given function =  = constant if    , . i.e., x  [1,  ). 4 2  Inverse Trigonometrical Functions 141 Example: 39 tan 1 x  cos 1 The number of positive integral solutions of the equation y 1y  sin 1 2 3 or 10 tan 1 x  cot 1 y  tan 1 3 is Solution: (b) (b) Two tan 1 x  tan 1 (c) Zero 1 1  tan 1 3 or tan 1 = tan 1 3  tan 1 x y y or tan 1 1 3x  tan 1 y 1  3x As x , y are positive integers, x  1 , 2 and corresponding y  2, 7  Solutions are (x, y)  (1, 2), (2, 7). Example: 40 ,  and  are three angles given by   2 tan 1 ( 2  1),   3 sin 1 1 Solution: (b,c)   2 tan 1 ( 2  1)  2 tan 1 tan   4  6  8 2  8   4 7 12    . Also,  cos 1 1 2 (d) None of these ID   3. (c)    (b)     1  1  sin 1    and   cos 1  . Then 3  2 E3 2 (a)    (d) None of these 1  3x  y 3x 60 (a) One 1 1 1 1     cos 1  cos 1 3 3 2 4 2 So,    U 1 Again cos 1   belongs to the first quadrant and  is in the second quadrant. 3   . a3 a  b3 b 1 1 co sec 2  tan 1   sec 2  tan 1  is equal to 2 b 2 a 2 2 D YG Example: 41 (b) (a  b) (a 2  b 2 ) (a) (a  b)(a 2  b 2 ) Soltuion: (c) Let tan 1 a b   , tan 1    b a (c) (a  b)(a 2  b 2 ) (d) None of these  tan   a , tan   b b a a a b b 1 1 cosec 2  tan 1   sec 2  tan 1  2 b 2 a 2 2 3 3 a3 b3 b3 a3   + = 1  cos  1  cos      1 2 sin 2   2 cos 2   2 2 U =  a 3 [ a 2  b 2  b] b 3 [ a 2  b 2  a]   a2  b 2    (a 2  b 2 )  b 2 (a 2  b 2 )  a 2    ST =  a 2  b 2 [a{ a 2  b 2  b}  b{ a 2  b 2  a}] = a3 b a b 2 b3  1 2 a a  b2 2 (rationalized) a 2  b 2 [ a 2  b 2 (a  b)]  (a 2  b 2 )(a  b). ST U D YG U ID E3 60 142 Inverse Trigonometrical Functions 141 Inverse Trigonometrical Functions 143 Properties of Inverse Trigonometrical Function The domain of sin 1 x is (a) ( ,  ) The range of tan 1 x is   (a)   ,   2 4.    (b)   ,   2 2  4 (b) – 1  2 1 3 5 3 (b) 0  3 U (b) (c) 10  3 [EAMCET 1985] (d) 1 [UPSEAT 2003] (d) 0 [EAMCET 2003] (c) 1 3 (d) (c) 2 3 (d) 2 4 9  6 2 , then x  3 ST If tan 1 x  2 cot 1 x  2 (b) 3 [Karnataka CET 1999] (c) 3 1 (d) 3 3 1 If 4 sin 1 x  cos 1 x   , then x is equal to (a) 0 10. (c) 2 D YG (b) (d) 1 The value of tan 1 x  2 cot 1 x is (a) 9. (d) (0,  )   1 1   1  cos cos 1    sin    7    7   (a) 8. (c) – 1 5  5    The value of cos 1  cos   sin 1  cos  is 3  3    (a)  7.  2 1 1  sin sin 1  cos 1   2 2  (a) 6. (c) ( ,  ) [Pb. CET 1997; DCE 2002] (b) (a) 0 5. [DCE 2002] sin 1 x  cos 1 x is equal to (a) (d) (, ) E3 3. (c) (0, 2 ) (b) [– 1, 1] ID 2. [Roorkee Screening 1993] U 1. 60 Basic Level (b) 1 2 [UPSEAT 2001] (c)  3 2 (d) 1 2 1 1  cos  2 cos 1  sin 1   5 5  (a) 2 6 5 [IIT 1981] (b)  2 6 5 (c) 1 5 (d)  1 5 144 Inverse Trigonometrical Functions 33    The value of sin 1  cos  is 5   (a) (c) (b) 1 (c) 5 (b) – 3 If sec 1 x  cosec –1 y, then cos 1 1 1  cos 1  x y (a)   4 (c) cos 2 (d)  cos 2   is   1 3 (c)   2 1 x 1 2  [Orissa JEE 2002] (d) 5 x2 1  2 [MNR 1978; MP PET 1989] 1  x2 (c) (d) x2  1 for some x  (1,1), then the value of cos 1 x is U If sin 1 x  (b) 1 3 (d)  1 If cos 1     , then tan   x (a) 18. (d) 30 o (c) D YG 17. (c) 15 o  3 The value of x which satisfies the equation tan 1 x  sin 1   10 (b) (d) None of these E3 1 (b)  (a) 3 16.  2 The value of cos(tan 1 (tan 2)) is 5 10 [MP PET 2003] (b) 90 o 1  (d)  [AMU 1988]  3   sin 1  1  is The value of sin 1   2  2   (a) 15.  10 60 1 2 (a) 45 o 14. 7 5 1 If sec 1    2 sin 1 (1)   , then x equals x (a) 13. (b) ID 12. 3 5 U 11. [DCE 1997; Karnataka CET 1996; IIT 1992] 3 10 ST (a) 19. (b) 5 10 (c) 7 10 (d) sec (cosec 1 x ) is equal to 9 10 [Kurukshetra CEE 2001] (a) cosec (sec 1 x ) 20. (d) None of these tan (cos 1 x ) is equal to (a) 21. (c)  (b) cot x 1 x2 x sin(cot 1 x )  2002] [IIT 1993] (b) x 1 x 2 (c) 1 x2 x (d) 1 x2 [MNR 1987; MP PET 2001; DCE Inverse Trigonometrical Functions 145 1  x2 (a) 22. (c) (1  x 2 )3 / 2 (b) x  (d) (1  x 2 ) cos(tan 1 x )  [MP 1 2 PET 1988; MNR 1981] 25 24 [Karnataka CET 1994] (b) 25 7 (c) (b)  2 (c) 1 2 (b) 5 3 sec 2 (tan 1 2)  cosec 2 (cot 1 3)  (b) 13 9 25 [EAMCET 1983] (d) 25 9 [EAMCET 2001] (c) 15 D YG (a) 5 (c) [Bihar CEE 1974] (d) None of these U 3 5 (d) None of these ID   1 3    sin tan 4    (a) 26. 24 25 The value of sin cot 1 tan cos 1 x is equal to (a) x 25. (d) None of these   7  cot cos 1     25   (a) 24. 1 x (c) 1  x 2 2 E3 23. 1 (b) 60 1 x2 (a) (d) 6 Advance Level 27. If cos 1 x  sin1 x , then 28. ST 1 2   , 2 2 (c) 0  x  1 2 (d)  1  x  1 (d)  1  x  1 2 (b)  1  x  2 2 2 (b)  3  3 , 8 8 (c)  3 7 3 32 , 8 (d) None of these If x satisfies the equation t 2  t  2  0, then there exists a value for (a) sin 1 x 31. 1 The greatest and the least values of (sin1 x )3  (cos 1 x )3 are (a) 30. (c) 0  x  If (cos 1 x )2  (sin 1 x )2  0, then (a) x  29. (b) 1  x  0 U (a) x  0 (b) cos 1 x (c) sec 1 x (d) None of these (c) x  (, 1]  [1,  ) (d) None of these If f (x )  sec 1 x  tan 1 x , then f (x ) is real for (a) x  [1, 1] (b) x  R 146 Inverse Trigonometrical Functions If  sin 1 x r  n  , then (a) n 7 ) 5 (c) (b) sin 1 (sin n(n  1) 2 (d) None of these 7 ) 5 (c) sec 1 (sec 7 ) 5 (d) None of these The number of real solutions of (x , y ) ; where | y |  sin x , y  cos 1 (cos x ),  2  x  2 is (c) 3 1 (d) 4 E3 (b) 1 The set of values of k for which x  kx  sin (sin 4 )  0 for all real x is 2 (a)  (b) (2, 2) (c) R  x2  1  1 x cos 1  x 2  1  x 2. 1   cos 1 x holds for   cos 2 4 2     (b) x  R (a) | x |  1 (d) None of these ID 36.  (b) 2n (a) 2 35. r 2 is the principal value of 5 (a) cos 1 (cos 34. x r 1 r 1 33. 2n 60 2n 32. (c) 0  x  1 (d) 1  x  0 U Sum and Difference of Inverse Trigonometrical Function 37. If cos 1 x  cos 1 y  cos 1 z  3 , then xy  yz  zx  2003] (a) 0 38. (b) 1 (b) 1 (b) tan 1   2 (d) – 3 [AMU 2001] (c) 1 6 (d) 1 7 ST [DCE 1999]  132  (c) tan 1    33  (d) None of these If sin 1 a  sin 1 b  sin 1 c   , then the value of a (1  a 2 )  b (1  b 2 )  c (1  c 2 ) will be (b) abc (c) 1 abc 2 [UPSEAT 1999] (d) 1 abc 3 If tan 1 x  tan 1 y  tan 1 A, then A  (b) x  y (a) x  y 42. (c) 3  1   2  tan 1    tan 1    11    12  (a) 2 abc 41. 7 6 U 5 6  33  (a) tan 1    132  40. [Karnataka CET 1 1  The value of tan  tan 1  tan 1  is 2 3  (a) 39. D YG Basic Level If tan 1 2 x  tan 1 3 x  2002]  4 then x = [MP PET 1988] (c) x y 1  xy (d) xy 1  xy [Roorkee 1978, 1980; MNR 1986; Karnataka CET Inverse Trigonometrical Functions 147 (a) –1   cot 1 (b)  3 (where x  y  0 )  (b) 4  6  4 1 2 (c) 3 4 4 2  tan cos 1  tan 1  = 5 3  1992] 6 17 (a) (d) None of these (b) 17 6 (d)  4 [EAMCET 1992] (d) None of these [IIT 1983; EAMCET 1988; MP PET 1990; MNR (c) 7 16 (d) 16 7 1 2 tan 1    tan 1   = 4   9 1 3 cos 1   2 5 (b) U (a) 50. (c) D YG (a)  49. (c) 3 2 (b) x x y  tan 1  y x y   1  [MP PET 1991; MNR 1990] 1 tan 1 (d) [AMU 1976, 1977] 2 48. 1    1 1  cos  tan 1  tan 1   3 2  (a) 47. (c) ID 46. [MP PET 1992] (b)     4 (d) 2 x , then x  3 3 8  tan 1  tan 1  4 5 19 tan 1 (a) (c) –1 60   cot 1 (a)    45. (d) None of these [AMU 1978] (b) 1 1 1 6 E3 If cot (c)  1, 3 4  sin 1  cos 1 x , then x  5 5 If cos 1 (a) 0 44. 1 6 U 43. (b) 1 sin 1 1 3 sin 1   2 5 [EAMCET 1994] 1 (c) tan 1   2 (d) Both (a) and (c)  cot 1 3 is equal to ST 5 [Karnataka CET 1995; MP PET 1993] (a) 51.  6  4 (c)  3 (d)  2 1   If sin sin 1  cos 1 x   1 , then x = 5   (a) 1 52. (b) (b) 0 A solution of the equation tan 1 (1  x )  tan 1 (1  x )  (a) x = 1 [UPSEAT 1994] (b) x = –1 (c)  2 4 5 (d) 1 5 is [Karnataka CET 1993] (c) x = 0 (d) x   148 Inverse Trigonometrical Functions (a) 3 3 , 4 8 If sin 1 (b) (c) 1 (d 3 [Karnataka CET 1994] (b)  2 4 (c) cos 1   5 E3  4  4 (b) 3 4 (c) If cos 1 x  cos 1 y  cos 1 z   then If tan 1 x  tan 1 y  tan 1 z  (b) x 2  y 2  z 2  2 xyz =0  2 , then (b) x  y  z  xyz  0 D YG (a) x  y  z  xyz  0 59. (d) None of these [EAMCET 1983] (b) 5 (a) x 2  y 2  z 2  xyz  0 58. 4 3 , 3 8 (d)  If tan 1 2, tan 1 3 are two angles of a triangle, then the third angle is (a) 57. (c) 3 1 sin 1    tan 1    5   7  (a) 56. 3 3 , 4 8 x 5   co sec 1    , then x  5 4 2 (a) 4 55. [ISM Dhanbad 1973] If tan 1 x  tan 1 y  tan 1 z   , then (a) 0  4 (d) None of these ID 54. x 1 2x  1 23  tan 1  tan 1 , then x = x 1 2x  1 36 60 If tan 1 (c) x 2  y 2  z 2  xyz  1 U 53. (c) xy  yz  zx  1  0 (d) x 2  y 2  z 2  2 xyz  1 [Karnataka CET 1996] (d) xy  yz  zx  1  0 1 1 1 =   xy yz zx (b) 1 [MP PET 1991] (c) 1 xyz (d) xyz Advance Level If we consider only the principal values of the inverse trigonometric functions, then the value of U 60. ST  1 tan  cos 1  sin 1  5 2  29 3 (a) 61. [IIT 1994] (b) 29 3 (c) 3 29 (d) The sum of first 10 terms of the series cot 1 3  cot 1 7  cot 1 13  cot 1 21 ....... is 5 (a) tan 1   6 62.   is (17 )  4 (b) tan 1 (100 ) 6 (c) tan 1   5 3 29 [Karnataka CET 1996]  1  (d) tan 1    100  3 3 3    Sum of infinite terms of the series cot 1 1 2    cot 1  2 2    cot 1 3 2    …….is 4 4 4    (a)  4 (b) tan 1 2 (c) tan 1 3 (d) None of these Inverse Trigonometrical Functions 149 tan 1 (a)  2 (b) (d) 0  4 (b) 60 (c) 4 (d) None of these    2 1      sin 1  3  2  ..............  sin 1  n  n  1  is  sin 1        2 6  12     n(n  1)  1  3 (c)  2 (d)   5 (b)  4 (c)  3 (d)  2 If sin 1 x  sin 1 y  sin 1 z  9 3 , then the value of x 100  y 100  z 100  101 is equal to 2 x  y 101  z 101 (a) 0 (b) 3 (c) – 3 (d) 9 U 68. 2 3 If sum of the infinite series cot 1 (2.12 )  cot 1 (2.2 2 )  cot 1 (2.2 3 )  cot(2.2 4 ) ..... is equal to (a) 67. (b) 2 The sum of the infinite series sin 1 (a) 66. (c) If sin 1 x  sin 1 y  sin1 z   , then x 4  y 4  z 4  4 x 2 y 2 z 2  K(x 2 y 2  y 2 z 2  z 2 x 2 ), where K = (a) 1 65.  4 [Karnataka CET 2000] E3 64. 1 1 1 1    tan 1  tan 1 .........  tan 1  2  ....... to  is equal 3 7 18  n  n 1  ID 63. If x1 , x 2 , x 3 , x 4 are roots of the equation x  x sin 2   x cos 2   x cos   sin   0, 4 3 2 (a)  69. If D YG then tan 1 x 1  tan 1 x 2  tan 1 x 3  tan 1 x 4  (b)   2 is a1, a2 , a3 ,........, an   d   d   tan 1  tan tan 1   1  a a 1  a a 1 2  2 3    (n  1) d a1  an (b) (n  1) d 1  a1an common difference (c) nd 1  a1 an (d) 3a  a3 1  3a2 d, Basic Level [MP PET 1993] (b) tan 1 3a  a3 1  3a2 (c) tan 1 If A  tan 1 x , then sin 2 A  3a  a3 1  3a2 (d) tan 1 3a  a3 1  3a2 [MNR 1988; UPSEAT 2000] (a) 2x 1x 2 then an  a1 an  a1 3 tan 1 a is equal to (a) tan 1 71. with Inverse Trigonometric Ratios of Multiple Angles ST 70. A.P.    d  ..........  tan 1     1  a a    n 1 n    U (a) an (d)   (c)    (b) 2x 1  x2 (c) 2x 1  x2 (d) None of these 150 Inverse Trigonometrical Functions sin(2 sin 1 0.8)  [MNR 1980] (a) 0.96 (a) Only 2 3 (b) Only   2 4 1  2 tan 1  5 3 (a)  2 (b) If sin 1 (c)  4 ab 1  ab  1 If 3 tan 1  2 3 (b)  2 (c) 0  3 (c)  4 (d)  4 (d) None of these b 1  ab (c) b 1  ab [MNR 1984] (d) (b) 2 [Pb. CET 2001; AMU 1992] (c) 3 (d) 1 10 3   =  U ST ab 1  ab (b) 2  x [UPSEAT 1986] (c)  2 (d)   4 1 sin cos 1   2 5  (a) [EAMCET 1983] [Dhanbad Engg. 1971]  1 1   tan 1  tan 1 , then x equals  x 3   1  sin x  1  sin x cot 1   1  sin x  1  sin x 2 2 nor  3 3 (d) None of these 2a 2b  sin 1  2 tan 1 x , then x = 2 1a 1  b2 (a)   x 80. (d) Neither [EAMCET 1981] (b) sin 1 (a) 1 79. 2 2 , 3 3 1 1 2 tan 1    tan 1    3   7  (a) 78. (c) D YG 77. 2 3  171  (b) cos 1    221   49  (a) tan 1    29  76. [Roorkee 1975]  15  1 cos 1    2 tan 1    17   5 (a) 75. 1 , then x  9 E3 74. (d) None of these 60 If cos( 2 sin 1 x )  (c) 0.64 ID 73. (b) 0.48 U 72. x 2 [Karnataka CET 2003] (b)  1 10 (c) 1 10 (d)  1 10 Advance Level 81.  ab  2 tan 1  tan   2   a  b [ISM Dhanbad 1976] Inverse Trigonometrical Functions 151  a cos   b  (a) cos 1    a  b cos   If cot 1[(cos  )1 / 2 ]  tan 1[(cos  )1 / 2 ]  x. Then sin x    (b) cot 2   2   x The value of sin 1  sin  2 3    x  k 2  kx  2 x 2  xk  k 2   (a) tan 1  2 2   x  2 xk  k  84. [AIEEE 2002]    1    cos cos 6     k   ,  where  x  2k , k  0  is 2 2 2   x  k  kx  x  x 2  2 xk  2k 2   x 2  2 xk  k 2   (c) tan 1  2 2   2 x  2 xk  2k   (b) tan 1  2 2   x  2 xk  k  Solution of equation sin[2 cos 1 {cot(2 tan 1 x )}]  0 is ID 1998; (b) A 3 5 2 D YG 1  5    tan  cos 1   3   2   (a) (c) C (b) 3 5 2 (c) ST U *** 2 3 5 Roorkee (d) All of these 1 3 The greater of the two angles A  2 tan 1 (2 2  1) and B  3 sin 1    sin 1   is 3 5 (a) B 86. (c) x  (1  2 ) only (b) x  1  2 only [IIT 1992] (d) None of these U 85. (d) None of these [UPSEAT 1992] (a) x  1 only   (d) cot   2 (c) tan  60   (a) tan 2   2 83.  b cos   (d) cos 1    a cos   b  E3 82.  a cos   (c) cos 1    a  b cos    a  b cos   (b) cos 1    a cos   b  [Roorkee 1986] (d) 2 3 5 ID E3 60 150 Inverse Trigonometrical Functions 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 b b b d a b c c b b d b d c a d d a a a 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 d b d a c c d d c c c b b c a c c d d a 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 c b b d a a b b d b d c d d a b d d b d 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 a b b b c b a b b d c a c d d a d b d a 81 82 83 84 85 86 a a c d b d ST U D YG 1 U Assignment (Basic & Advance Level)