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If tan^-1 x – tan^-1 y = tan^-1 A, then A =

1) x - y 2) x + y 3) [x - y] / [1 + xy] 4) [x + y] / [1 - xy] Solution: (3) [x - y] / [1 + xy] tan-1 x - tan-1 y = tan-1 A tan-1... View Article

Let cos (2 tan^-1 x) = 1 / 2, then the value of x will be

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

If tan^-1 x + tan^-1 y + tan^-1 z = ? / 2, then

1) x + y + z - xyz = 0 2) x + y + z + xyz = 0 3) xy + yz + zx + 1 = 0 4) xy + yz + zx - 1 = 0 Solution: (4) xy + yz + zx - 1 = 0... View Article

cot^-1 (3 / 4) + sin^-1 (5 / 13) =

1) sin-1 63 / 65 2) sin-1 12 / 13 3) sin-1 65 / 68 4) sin-1 5 / 12 Solution: (1) sin-1 63 / 65 Let cot-1 3 / 4 = θ cot θ = 3 / 4... View Article

tan^-1 1 + tan^-1 2 + tan^-1 3 =

1) π / 2 2) π / 3 3) 0 4) None of these Solution: (4) None of these tan-1 1 + tan-1 2 + tan-1 3 = tan-1 1 + π + tan-1 (5 / - 5)... View Article

tan [cos^-1 (4 / 5) + tan^-1 (2 / 3)] =

1) 6 / 17 2) 17 / 6 3) 7 / 16 4) 16 / 7 Solution: (2) 17 / 6 tan [cos-1 (4 / 5) + tan-1 (2 / 3)] = tan [tan-1 √1 - (16 / 25)] /... View Article

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

1) 90o 2) 60o 3) 45o 4) tan-1 2 Solution: (4) tan-1 2 2 tan-1 (1 / 3) + tan-1 (1 / 2) = tan-1 [(2 / 3)] / [1 - (1 / 9)] + tan-1... View Article

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

1) 1 2) 2 3) 3 4) 4 Solution: (1) 1 sin [sin-1 (1 / 3) + sec-1 3] + cos (tan-1 (1 / 2) + tan-1 2) = sin [sin-1 (1 / 3) + cos-1 (1... View Article

cos^-1 (4 / 5) + tan^-1 (3 / 5) =

1) tan-1 (27 / 11) 2) sin-1 (11 / 27) 3) cos-1 (11 / 27) 4) None of these Solution: (1) tan-1 (27 / 11) cos-1 (4 / 5) + tan-1 (3... View Article

If sin [cot^-1 (x + 1)] = cos (tan^-1 x), then x =

1) - 1 / 2 2) 1 / 2 3) 0 4) 9 / 4 Solution: (1) - 1 / 2 sin [cot-1 (x + 1)] = sin [sin-1 1 / (√x2 + 2x + 2)] = 1 / (√x2 + 2x + 2)... View Article

sin [cot^-1 (cos tan^-1 x)] =

1) x / (√x2 + 2) 2) x / (√x2 + 1) 3) 1 / (√x2 + 2) 4) √(x2 + 1) / (x2 + 2) Solution: (4) √(x2 + 1) / (x2 + 2) sin [cot-1 (cos... View Article

sin [(? / 2) – sin^-1 (- ?3 / 2)] =

1) √3 / 2 2) - √3 / 2 3) 1 / 2 4) - 1 / 2 Solution: (3) 1 / 2 sin [(π / 2) - sin-1 (- √3 / 2)] = cos sin-1 (- √3 / 2) = cos... View Article

If sin^-1 x + sin^-1 y + sin^-1 z = ? / 2, then the value of x² + y² + z² + 2xyz =

1) 0 2) 1 3) 2 4) 3 Solution: (2) 1 Let sin-1 x = α, sin-1 y = β, sin-1 z = γ. x = sin α, y = sin β, z = sin γ sin-1 x + sin-1 y... View Article

The value of cos {tan^-1 (tan 2)} is

1) 1 / √5 2) - 1 / √5 3) cos 2 4) - cos 2 Solution: (4) - cos 2 cos {tan-1 (tan 2)} = cos (tan-1 - (tan (π - 2))) = cos (- π - 2) = cos... View Article

sin^-1 (?x) / (?x + a) =

1) cos-1 (√x / a) 2) cosec-1 (√x / a) 3) tan-1 (√x / a) 4) None of these Solution: (3) tan-1 (√x / a) Put x = a tan2 θ sin-1 (√x)... View Article

The value of sin cot^-1 tan cos^-1 x =

1) x 2) π / 2 3) 1 4) None of these Solution: (1) x Let cos-1 x = θ x = cos θ sec θ = 1 / x tan θ = √sec2 θ - 1 = √(1 / x2) - 1... View Article

cos^-1 (cos (7? / 6)) =

1) 7π / 6 2) 5π / 6 3) π / 6 4) None of these Solution: (2) 5π / 6 cos-1 (cos (7π / 6)) = cos-1 (cos (π + (π / 6))) = cos-1 (cos... View Article

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

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

sin^-1 [x ?1 – x – ?x ?1 – x²] =

1) sin-1 x + sin-1 √x 2) sin-1 x - sin-1 √x 3) sin-1 √x - sin-1 x 4) None of these Solution: (2) sin-1 x - sin-1 √x Let x = sin θ... View Article

If cos^-1 x = ?, (0 < x < 1) and sin^-1 (2x ?1 – x²) + sec^-1 1 / (2x² – 1) = 2? / 3, then tan^-1 2x =

1) π / 6 2) π / 4 3) π / 3 4) π / 2 Solution: (3) π / 3 cos-1 x = ɑ x = cos ɑ sin-1 (2 cos ɑ √1 - cos2 ɑ) + sec-1 1 / (2cos2 ɑ - 1) = 2π /... View Article

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