Question

$\tan [\frac{1}{2}si{n}^{-1}\frac{2x}{1+x^2}+\frac{1}{2}{\cos }^{-1}\frac{1-y^2}{1+y^2}]$ is equal to

• A. $x+y$
• B. $x-y$
• C. $\frac{x+y}{1-xy}$
• D. $\frac{x-y}{1+xy}$

Answer 1

The correct option is C $\frac{x+y}{1-xy}$

Given, $\tan \left[\frac{1}{2}{\sin }^{-1}\right(\frac{2x}{1+x^2})+\frac{1}{2}{\cos }^{-1}(\frac{1-y^2}{1+y^2}\left)\right]$
$=\tan \left[\frac{2ta{n}^{-1}\left(x\right)+2{\tan }^{-1}\left(y\right)}{2}\right]$
$=\tan \left[{\tan }^{-1}\right(x)+{\tan }^{-1}(y\left)\right]$
$=\tan \left[{\tan }^{-1}\right(\frac{x+y}{1-xy}\left)\right]$
$=\frac{x+y}{1-xy}$