Question

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

• A. $\frac{2\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\pi$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is C $\pi$

For $x>1$
$2{\tan }^{-1}x=\pi -{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\therefore 2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$=\pi -{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\pi$