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If a₁, a₂, a₃, …. an are in AP with common difference 5 and if a_ia_j ≠ – 1 for i, j = 1, 2, …. N, then tan^-1 5 / [1 + a₁a₂] + tan^-1 5 / [1 + a₂a₃] + …. + tan^-1 5 / [[1 + a_n-1a_n] =

1) tan-1 5 / [1 + an-1an] 2) tan-1 5a1 / [1 + ana1] 3) tan-1 [5n - 5] / [1 + ana1] 4) tan-1 [5n - 5] / [1 + an+1a1] 5) tan-1 5n / [1... View Article

The value of sin^-1 |cos (4095^o)|is

1) -(π/3) 2) π/6 3) -(π/4) 4) π/4 5) π/2 Solution: (3) - (π / 4) sin-1{cos (4095o)} = sin-1{cos (11 . 360o + 135o)} = sin-1... View Article

The value of sin [tan^-1 [1 – x²] / 2x + cos^-1 [1 – x²] / [1 + x²] is

1) 1 2) 0 3) - 1 4) π/2 Solution: (1) 1 Put x = tan θ sin [tan-1 [1 - x2] / 2x + cos-1 [1 - x2] / [1 + x2] = sin [tan-1 (cot 2θ)... 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 θ. Then θ is equal

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

If a sin^-1 x – b cos^-1 x = c, than a sin^-1 x + b cos^-1 x is equal to

1) [πab + c (a - b)] / (a + b) 2) 0 3) [πab - c (a - b)] / (a + b) 4) π / 2 Solution: (1) [πab + c (a - b)] / (a + b) a sin-1 x -... View Article

If tan^-1 x = π / 4 – tan^-1 (1/3), then x is equal to

1) 1/3 2) 1/2 3) 1/4 4) 1/6 Solution: (2) ½ tan-1 x = π / 4 - tan-1 (1/3) tan-1 x + tan-1 (1 / 3) = π / 4 Clearly x = [3 - 1] /... View Article

If sin^-1 x + sin^-1 y + sin^-1 z = 3π / 2, then the value of x⁹ + y⁹ + z⁹ – 1 / x⁹ y⁹ z⁹ is

1) 0 2) 1 3) 2 4) 3 Solution: (3) 2 sin-1 x + sin-1 y + sin-1 z = 3π / 2 sin-1 x = sin-1 y = sin-1 z = π / 2 x = y = z = 1 x9 +... View Article

If tan^-1 (tan (5π / 4)) = α, tan^-1 (- tan (2π / 3)) = β, then

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

The value of 4 tan^-1 1 / 5 – tan^-1 1 / 70 + tan^-1 1 / 99 is

1) π / 4 2) π / 2 3) π 4) ∞ Solution: (1) π / 4 4 tan-1 1 / 5 - tan-1 1 / 70 + tan-1 1 / 99 = 2{2tan-1 (1 / 5)} - {tan-1 1 / 70 -... View Article

The value of cot {∑_n=1²³ cot^-1 (1 + ∑_k=1ⁿ 2k)} is

1) 23/25 2) 25/23 3) 23/24 4) 24/23 Solution: (2) 25/23 cot {∑n=123 cot-1 (1 + ∑k=1n 2k)} = cot {∑n=123 cot-1 (1 + 2 * [n (n +... View Article

The number of real solutions of tan^-1 √x (x + 1) + sin^-1 √x² + x + 1 = π / 2 is

1) zero 2) one 3) two 4) infinite Solution: (1) zero The given function is tan-1 √x (x + 1) + sin-1 √x2 + x + 1 = π / 2 Function... View Article

The value of sin (2 tan^-1 x), |x| ≤ 1 is

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

The number of value(s) of x, in which tan^-1 (x / [x + 2]) + (π / 2) = tan^-1 (2x²) + cot^-1 (x / [x + 4]), 0 < x < 1

1) not defined 2) √3 3) ± 1 4) 0 Solution: (4) 0 tan-1 (x / [x + 2]) + (π / 2) = tan-1 (2x2) + cot-1 (x / [x + 4]) tan-1 (x / [x... View Article

tan^-1 (√3) – cot^-1 (-√3) is equal to

1) 0 2) 2√3 3) - (π / 2) 4) π Solution: (3) - (π / 2) tan-1 (√3) - cot-1 (-√3) = cot-1 (- √3) = π - cot-1 (- √3) = tan-1 (√3) -... View Article

If we consider only the principle values of the inverse trigonometric functions then the value of tan [cos^-1 (1 / 5√2) – sin^-1 (4 / √17)] is

1) √29 / 3 2) 29 / 3 3) √3 / 29 4) 3 / 29 Solution: (4) 3 / 29 tan (cos-1 (1 / 5 √2) - sin-1 (4 / √17)) = tan (π / 2 - sin-1 (1 /... View Article

[latex]\frac{d}{dx}\left [ \left ( \frac{\tan ^{2}2x-\tan ^{2}x}{1-\tan ^{2}2x\tan ^{2}x} \right ) \cot 3x\right ][/latex]

(1) sec2 x (2) sec x tan x (3) tan 2x tan x (4) none of these Solution: Let y = ((tan2 2x - tan2 x)/(1 - tan2 2x tan2 x)) cot 3x... View Article

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