If $y = 2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$, then …………….$< y <$…..…………..

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

$therefore,$ $y = 2{\tan ^{ - 1}}\tan \theta + {\sin ^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$ [let $x = \tan \theta ]$

$\Rightarrow$ $y = 2\theta + {\sin ^{ - 1}}\sin 2\theta$

$\Rightarrow$ $y = 2\theta + 2\theta = 4\theta$
$\Rightarrow$ $y = 4{\tan ^{ - 1}}x$
$- \pi /2 < {\tan ^{ - 1}}x < \pi /2$

$therefore,$ $- \frac{{4\pi }}{2} < 4{\tan ^{ - 1}}x < 4\pi /2$

$\Rightarrow$ $- 2\pi < 4{\tan ^{ - 1}}x < 2\pi$
$\Rightarrow$ $- 2\pi < y < 2\pi$