Question:
If $x>1$, then $2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ is equal to
(a) $4 \tan ^{-1} x$
(b) 0
(c) $\frac{\pi}{2}$
(d) $\pi$
Solution:
$2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x+2 \tan ^{-1} x \quad\left[\because \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=2 \tan ^{-1} x\right]$
$=4 \tan ^{-1} x$
Hence, the correct answer is option (a)