If ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$, where $\left. {a,x \in } \right]0,1$

We have, ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$

Let $a = \tan \theta \Rightarrow \theta = {\tan ^{ - 1}}a$

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

$\Rightarrow$ ${\sin ^{ - 1}}\sin 2\theta + {\cos ^{ - 1}}\cos 2\theta = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\Rightarrow$ $2\theta + 2\theta = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\Rightarrow$ $4{\tan ^{ - 1}}a = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\Rightarrow$ $2 \cdot 2{\tan ^{ - 1}}a = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\Rightarrow$ $2 \cdot {\tan ^{ - 1}}\frac{{2a}}{{1 - {a^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{2 \cdot \left( {\frac{{2a}}{{1 - {a^2}}}} \right)}}{{1 - {{\left( {\frac{{2a}}{{1 - {a^2}}}} \right)}^2}}} = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$
$therefore, x = \frac{{2a}}{{1 - {a^2}}}$