If sin^-1(2a/(1 + a^2)) + cos^-1(1-a^2)/(1 + a^2) = tan^-1(2x)/1- x^2, where a, x ∈ (0, 1) then, the value of x is A. 0 B. a/2 C. a D. 2a/(1 - a^2)

Question

If sin^-1\((\frac{2a}{1+a^2})\) + cos^-1\((\frac{1-a^2}{1+a^2})\) = tan^-1\((\frac{2x}{1-x^2})\), where a, x ∈ (0, 1) then, the value of x is

A. 0

B. \(\frac{a}2\)

C. a

D. \(\frac{2a}{1-a^2}\)

Answer 1

Correct answer is D.\(\frac{2a}{1-a^2}\) 

We are given that,

 sin^-1\((\frac{2a}{1+a^2})\) + cos^-1\((\frac{1-a^2}{1+a^2})\) = tan^-1\((\frac{2x}{1-x^2})\)

Where, a, x ∈ (0, 1).

We need to find the value of x.

Using property of inverse trigonometry,

Taking tangent on both sides,