$\tan \frac{1}{2}\left[ {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right],\;\,|x|\; < 1,\;y > 0$ and $xy < 1$

Putting $x = \tan \theta$ and $y = \tan \phi ,$ we get
$\tan \left\{ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right\}$

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

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

$= \tan \left\{ {\frac{1}{2} \times 2\theta + \frac{1}{2} \times 2\phi } \right\} = \tan (\theta + \phi ) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{x + y}}{{1 - xy}}$