Question

If $x>1$, then $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ is equal to

• A. $4{\tan }^{-1}x$
• B. $0$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

Answer: (A)

$4{\tan }^{-1}x$

Given, $2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Let $x=\tan \theta$

$=2{\tan }^{-1}\tan \theta +{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$

$=2\theta +{\sin }^{-1}\left(\sin 2\theta \right)$

$=2\theta +2\theta$

$=4\theta$

$=4{\tan }^{-1}x$ $[\because x=\tan \theta ]$