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The value of $\tan \left\{\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right\}$ is
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Verified Answer
The correct answer is:
$\frac{x+y}{1-x y}$
Let $x=\tan \theta$ and $y=\tan \phi$
$\begin{aligned}
& \Rightarrow \tan \left\{\frac{1}{2} \sin ^{-1} \sin 2 \theta+\frac{1}{2} \cos ^{-1} \cos 2 \phi\right\} \\
& =\tan (\theta+\phi) \\
& =\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}=\frac{x+y}{1-x y}
\end{aligned}$
$\begin{aligned}
& \Rightarrow \tan \left\{\frac{1}{2} \sin ^{-1} \sin 2 \theta+\frac{1}{2} \cos ^{-1} \cos 2 \phi\right\} \\
& =\tan (\theta+\phi) \\
& =\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}=\frac{x+y}{1-x y}
\end{aligned}$
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