Putting $a = \tan \theta $ and $b = \tan \phi $
So, ${\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + {\sin ^{ - 1}}\left( {\frac{{2\tan \phi }}{{1 + {{\tan }^2}\phi }}} \right) = 2{\tan ^{ - 1}}x$
==> ${\sin ^{ - 1}}\sin (2\theta ) + {\sin ^{ - 1}}\sin (2\phi ) = 2{\tan ^{ - 1}}x$
==> $2(\theta + \phi ) = 2{\tan ^{ - 1}}x$
Hence $x = \tan (\theta + \phi )$
==> $x = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$
Substituting these values, we get $x = \frac{{a + b}}{{1 - ab}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.