If $x\in (1,\mathrm{\infty }),$ then ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ equals

Solution:

Let ${\tan }^{-1}x=\theta \text{.}$ Then, $x=\tan \theta$

Now,

       $x\in (1,\mathrm{\infty })\Rightarrow \tan \theta >1\Rightarrow \frac{\pi }{4}<\theta <\frac{\pi }{2}\Rightarrow \frac{\pi }{2}<2\theta <\pi$

$\begin{array}{l}\therefore {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\\ = {\sin }^{-1}(\sin 2\theta )\\ = {\sin }^{-1}(\sin (\pi -2\theta \left)\right)\end{array}$

     $= \pi -2\theta \left[\because \frac{\pi }{4}<\theta <\frac{\pi }{2}\Rightarrow -\frac{\pi }{2}<\pi -2\theta <0\right]$

      $=\pi -2{\tan }^{-1}x$