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. $\frac{2a}{1+a^2}$
• B. 0
• C. $\frac{2a}{1-a^2}$
• D. $\frac{a}{2}$

Answer 1

Answer: (C)

$\frac{2a}{1-a^2}$

Answer 2

The correct answer is (C) : $\frac{2a}{1-a^2}$.