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Question

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

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

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

0

$\frac{a}{2}$

Solution

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