The value of $tan\frac{1}{2}\begin{Bmatrix}sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}\end{Bmatrix}$ if $|x|<1. y > 0 $ and $xy < 1$, is:

$\frac{x+y}{1-xy}$

The correct answer is Option (3) → $\frac{x+y}{1-xy}$

$\tan\frac{1}{2}\left\{\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right\}$

as $\sin^{-1}\frac{2x}{1+x^2}=θ$

$\tan^{-1}\frac{2x}{1+x^2}=θ$

$\cos^{-1}\frac{1-y^2}{1+y^2}=\phi$

$\tan^{-1}\frac{2y}{1+y^2}=\phi⇒\tan\frac{1}{2}\left(\tan^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right)$

$=\tan\frac{1}{2}(2\tan^{-1}x+2\tan^{-1}y)$

$=\frac{x+y}{1-xy}$