The solution of the equation sin-1 (tan ? / 4) – sin-1 (?3 / x) – (x / 6) = 0 is given by x = 1) 1 2) 2 3) 3 4) 4 Solution: (4) 4 The given equation is sin-1 (tan Ï€ / 4) - sin-1 (√3 / x) - (x / 6) = 0 It becomes sin-1 1 - sin-1 √3... View Article
If tan-1 1 / [1 + 2] + tan-1 1 / [1 + 2 – 3] + tan-1 1 / [1 + 3 – 4] + …. + tan-1 1 / [1 + n (n + 1)] = tan-1 ?, then ? = 1) n 2) n / (n + 1) 3) n / (n + 2) 4) n + 1 Solution: (3) n / (n + 2) LHS = tan-1 1 / [1 + 2] + tan-1 1 / [1 + 2 - 3] + tan-1 1 /... View Article
If ?C = 90o in triangle ABC, then tan-1 b / (b + c) + tan-1 b / (c + a) = 1) π / 2 2) π / 4 3) π / 3 4) π Solution: (2) π / 4 tan-1 b / (b + c) + tan-1 b / (c + a) = tan-1 [a / (b + c)] + [b / (c + a)] /... View Article
tan [(1 / 2) sin-1 2x / [1 + x2] + (1 / 2) cos-1 [1 – y2] / [1 + y2] where (xy ? 1) = 1) x + y 2) [x + y] / [1 - xy] 3) [x + y] / [1 + xy] 4) [x - y] / [1 + xy] Solution: (2) [x + y] / [1 - xy] tan [(1 / 2) sin-1 2x... View Article
tan-1 x + tan-1 y + tan-1 [1 – x – y + xy] / [1 + x + y – xy] = 1) Ï€ / 2 2) Ï€ / 4 3) Ï€ 4) Ï€ / 3 Solution: (2) Ï€ / 4 LHS = tan-1 x + tan-1 y + tan-1 [1 - x - y + xy] / [1 + x + y - xy] = tan-1... View Article
If sin-1 x + sin-1 (1 – x) = sin-1 ?1 – x2, then x = 1) 0, 1 / 2 2) 1, - 1 / 2 3) - 1 / 2, 1 / 2 4) none of these Solution: (1) 0, 1 / 2 Put x = sin θ sin-1 (sin θ) + sin-1 (1 - sin... View Article
If sin-1 x – cos-1 x = ? / 6, then x = 1) 1/2 2) √3/2 3) -1/2 4) None of these Solution: (2) √3 / 2 sin-1 x - cos-1 x = Ï€ / 6 (Ï€ / 2) - cos-1 x - cos-1 x = Ï€ / 6 2... View Article
The principal value of sin-1 [sin (2? / 3)] is 1) - 2π / 3 2) 2π / 3 3) 4π / 3 4) None of these Solution: (4) None of these Let sin-1 [sin (2π / 3)] = θ sin θ = sin (2π / 3)... View Article
If tan-1 a + tan-1 b = sin-1 1 – tan-1 c, then 1) a + b + c = abc 2) ab + bc + ca = abc 3) 1/a + 1/b + 1/c - (1/abc) = 0 4) ab + bc + ca = a + b + c Solution: (3) 1/a + 1/b + 1/c -... View Article
The number of positive integral solutions of the equation tan-1 x + cos-1 (y) / (?1 + y2) = sin-1 (3 / ?10) is 1) one 2) two 3) zero 4) None of these Solution: (2) two tan-1 x + cos-1 (y) / (√1 + y2) = sin-1 (3 / √10) tan-1 x + tan-1 (1 /... View Article
For the equations cos-1 x + cos-1 2x + ? = 0, the number of real solutions is 1) 1 2) 2 3) 0 4) ∞ Solution: (3) 0 cos-1 x + cos-1 2x + π = 0 cos-1 x + cos-1 2x = - π cos-1 2x = - π - cos-1 x 2x = cos (π +... View Article
If x, y and z are in AP and tan-1 x, tan-1 y and tan-1 z are also in AP, then 1) x = y = z 2) x = y = - z 3) x = 1, y = 2, z = 3 4) x = 2, y = 4, z = 6 5) x = 2, y = 3z Solution: (1) x = y = z Given x, y,... View Article
The value of tan [cos-1 (4 / 5) + tan-1 (2 / 3)] is 1) 6 / 17 2) 17 / 6 3) 16 / 7 4) none Solution: (2) 17 / 6 tan [cos-1 (4 / 5) + tan-1 (2 / 3)] = tan [tan-1 (3 / 4) + tan-1 (2 /... View Article
If sin [sin-1 (1 / 5) + cos-1 x] = 1, then the value of x is 1) - 1 2) 2/5 3) 1/3 4) 1 5) 1/5 Solution: (5) 1 / 5 sin [sin-1 (1 / 5) + cos-1 x] = 1 (sin-1 (1 / 5) + cos-1 x) = sin-1 1... View Article
Which one of the following is correct ? 1) sin (cos-1 x) = cos (sin-1 x) 2) sec (tan-1 x) = tan (sec-1 x) 3) cos (tan-1 x) = tan (cos-1 x) 4) tan (sin-1 x) = sin (tan-1 x)... View Article
If 4 sin-1 x + cos-1 x = ?, then x is equal to 1) 0 2) 1/2 3) - (1/2) 4) 1 Solution: (2) 1/2 4 sin-1 x + cos-1 x = π 4 sin-1 x + [(π / 2) - sin-1 x] = π 4 sin-1 x - sin-1 x =... View Article
tan-1 [3x – x3] / [1 – 3x2] – tan-1 [2x] / [1 – x2] = 1) 0 2) 1 3) tan-1 x 4) tan-1 2x Solution: (3) tan-1 x tan-1 [3x - x3] / [1 - 3x2] - tan-1 [2x] / [1 - x2] tan-1 a - tan-1 b =... View Article
cot-1 (?cos ?) – tan-1 (?cos ?) = x, then sinx is 1) tan2 (α/2) 2) cot2 (α/2) 3) tanα 4) cot (α/2) Solution: (1) tan2 (α/2) Put √cos α = θ cot-1 θ - tan-1 θ = x (Ï€ / 2) - 2 tan-1... View Article
The equation sin-1 x – cos-1 x = cos-1 (?3 / 2) has 1) no solution 2) unique solution 3) infinite number of solutions 4) None of the above Solution: (2) unique solution sin-1 x -... View Article
If tan-1 [1 – x] / [1 + x] = 1 / 2 tan-1 x, then the value of x is 1) 1/2 2) 1/√3 3) √3 4) 2 Solution: (2) 1 / √3 Put (1 / 2) tan-1 x = θ x = tan 2θ [1 - x] / [1 + x] = [1 - tan 2θ] / [1 + tan... View Article
