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If $\sin^{- 1} (\frac{2 a}{1 + a^{2}}) + \sin^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 \tan^{- 1} x,$ then $x,$ is equal to

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Correct option is D

We have, sin−1⁡2x1+x2=2tan−1⁡x sin−1⁡2a1+a2+sin−1⁡2b1+b2=2tan−1⁡x⇒ 2tan−1a+2tan−1⁡b=2tan−1⁡x= tan−1⁡a+tan−1⁡b=tan−1⁡x= tan−1⁡a+b1−ab=tan−1⁡x⇒x=a+b1−ab

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If sin−1⁡2a1+a2+sin−1⁡2b1+b2=2tan−1⁡x, then x, is equal to

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If $\sin^{- 1} (\frac{2 a}{1 + a^{2}}) + \sin^{- 1} (\frac{2 b}{1 + b^{2}}) = 2 \tan^{- 1} x,$ then $x,$ is equal to