if the
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)

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