Question:
If $x>1$, then write the value of $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ in terms of $\tan ^{-1} x$.
Solution:
$\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi-2 \tan ^{-1} x$
$\left[\because 2 \tan ^{-1} x=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right.$ for $\left.x>1\right]$
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