Archives

The value of sin [2 tan^-1 (1 / 3)] + cos (tan^-1 2?2) =

1) 16 / 15 2) 14 / 15 3) 12 / 15 4) 11 / 15 Solution: (2) 14 / 15 sin [2 tan-1 (1 / 3)] + cos (tan-1 2√2) sin [2 tan-1 (1 / 3) ]... View Article

If 3 sin^-1 [2x] / [1 – x²] – 4 cos^-1 [1 – x²] / [1 + x²] + 2 tan^-1 [2x] / [1 – x²] = ? / 3, then x =

1) √3 2) 1 / √3 3) 1 4) None of these Solution: (2) 1 / √3 3 sin-1 [2x] / [1 - x2] - 4 cos-1 [1 - x2] / [1 + x2] + 2 tan-1 [2x] /... View Article

The value of cos^-1 (cos 12) – sin^-1 (sin 14) is

1) - 2 2) 8π - 26 3) 4π + 2 4) None of these Solution: (2) 8π - 26 cos-1 (cos 12) - sin-1 (sin 14) = 4π - 12 - (14 - 4π) = 8π - 26... View Article

If tan^-1 x + tan^-1 y = ? / 4, then

1) x + y + xy = 1 2) x + y - xy = 1 3) x + y + xy + 1 = 0 4) x + y - xy + 1 = 0 Solution: (1) x + y + xy = 1 Let tan-1 x + tan-1... View Article

4 tan^-1 (1 / 5) – tan^-1 (1 / 239) =

1) π 2) π / 2 3) π / 3 4) π / 4 Solution: (4) π / 4 2 tan-1 x = tan-1 [2x] / [1 - x2] 4 tan-1 (1 / 5) = 2 [2 tan-1 (1 / 5)] = 2... View Article

3 tan^-1 a =

1) tan-1 [3a + a3] / [1 + 3a2] 2) tan-1 [3a - a3] / [1 + 3a2] 3) tan-1 [3a + a3] / [1 - 3a2] 4) tan-1 [3a - a3] / [1 - 3a2] Solution:... View Article

sin (4 tan^-1 1 / 3) =

1) 12 / 25 2) 24 / 25 3) 1 / 5 4) None of these Solution: (2) 24 / 25 sin (4 tan-1 1 / 3) = sin [2 tan-1 (2 / 3) / (1 - (1 / 9))]... View Article

(1 / 2) cos^-1 [1 – x] / [1 + x] =

1) cot-1 √x 2) tan-1 √x 3) tan-1 x 4) cot-1 x Solution: (2) tan-1 √x Let x = tan2 θ θ = tan-1 √x (1 / 2) cos-1 [1 - x] / [1 + x]... View Article

tan [(1 / 2) cos^-1 (?5 / 3)] =

(a) [3 - √5] / 2 (b) [3 + √5] / 2 (c) 2 / [3 - √5] (d) 2 / [3 + √5] 1) a and d 2) d only 3) b only 4) None of these... View Article

If 2 cos^-1 ?(1 + x) / 2 = ? / 2, then x =

1) 1 2) 0 3) -1/2 4) 1/2 Solution: (2) 0 2 cos-1 √(1 + x) / 2 = π / 2 cos-1 √(1 + x) / 2 = π / 4 cos (π / 4) = √(1 + x) / 2 (1... View Article

If sin^-1 x + cot^-1 (1 / 2) = ? / 2, then x is

1) 0 2) 1 / √5 3) 2 / √5 4) √3 / 2 Solution: (2) 1 / √5 sin-1 x + cot-1 (1 / 2) = π / 2 sin-1 x = (π / 2) - cot-1 (1 / 2) =... View Article

If tan^-1 x + 2 cot^-1 x = 2? / 3, then x =

1) √2 2) 3 3) √3 4) [√3 - 1] / [√3 + 1] Solution: (3) √3 tan-1 x + 2 cot-1 x = 2π / 3 tan-1 x + 2 [(π / 2) - tan-1 x] = 2π / 3 -... View Article

The solution of sin^-1 x – sin^-1 2x = ± (? / 3) is

1) ± 1 / 3 2) ± 1 / 4 3) ± √3 / 2 4) ± 1 / 2 Solution: (4) ± 1 / 2 sin-1 x - sin-1 2x = ± (π / 3) sin-1 2x = sin-1 x - sin-1 (√3... View Article

tan^-1 (1 / 2) + tan^-1 (1 / 3) =

1) 0 2) π / 4 3) π / 2 4) π Solution: (2) π / 4 tan-1 (1 / 2) + tan-1 (1 / 3) = tan-1 [(1 / 2) + (1 / 3)] / [1 - (1 / 2) (1 /... View Article

For the equation cos^-1 x + cos^-1 2x + ? = 0, the number of real solution is

1) 1 2) 2 3) 0 4) Infinity 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 =... View Article

sin [3 sin^-1 (1 / 5)] =

1) 71 / 125 2) 74 / 125 3) 3 / 5 4) 1 / 2 Solution: (1) 71 / 125 sin [3 sin-1 (1 / 5)] = sin [sin-1 {3 * (1 / 5) - 4 (1 / 5)3}]... View Article

tan^-1 [c₁x – y] / [c₁x + x] + tan^-1 [c₂ – c₁] / [1 + c₂c₁] + tan^-1 [c₃ – c₂] / [1 + c₃c₂] ….. + tan^-1 (1 / c_n) =

1) tan-1 (y / x) 2) tan-1 yx 3) tan-1 (x / y) 4) tan-1 (x - y) Solution: (3) tan-1 (x / y) tan-1 [c1x - y] / [c1x + x] + tan-1 [c2... View Article

If tan^-1 [x – 1] / [x + 1] + tan^-1 [2x – 1] / [2x + 1] = tan^-1 (23 / 36), then x =

1) (3 / 4), (- 3 / 8) 2) (3 / 4, 3 / 8) 3) (4 / 3, 3 / 8) 4) None of these Solution: (4) None of these tan-1 [x - 1] / [x + 1] + tan-1 [2x -... View Article

If tan (x + y) = 33 and x = tan^-1 3, then y will be

1) 0.3 2) tan-1 1.3 3) tan-1 (0.3) 4) tan-1 (1 / 18) Solution: (3) tan-1 (0.3) tan (x + y) = 33 x + y = tan-1 33 y = tan-1 33 -... View Article

If (tan^-1 x)² + (cot^-1 x²) = 5?² / 8, then x =

1) -1 2) 1 3) 0 4) None of these Solution: (1) -1 (tan-1 x)2 + (cot-1 x2) = 5π2 / 8 (tan-1 x)2 + [(π / 2) - tan-1 x]2 = 5π2 / 8... View Article