If $|x| \le 1$, then $2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is equal to

We have, $2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}$

Let $x = \tan \theta$
$therefore, 2{\tan ^{ - 1}}\tan \theta + {\sin ^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$

$= 2\theta + {\sin ^{ - 1}}\sin 2\theta$
$= 2\theta + 2\theta$
$= 4\theta$
$= 4{\tan ^{ - 1}}x$