The value of $\tan \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$ is

We have, $\tan \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right)$

Let $\frac{1}{2}{\cos ^{ - 1}}\frac{2}{{\sqrt 5 }} = \theta$
$\Rightarrow$ ${\cos ^{ - 1}}\frac{2}{{\sqrt 5 }} = 2\theta \Rightarrow \cos 2\theta = \frac{2}{{\sqrt 5 }}$

$therefore, \left( {1 - 2{{\sin }^2}\theta } \right) = \frac{2}{{\sqrt 5 }}$

$\Rightarrow$ $2{\sin ^2}\theta = 1 - \frac{2}{{\sqrt 5 }}$

$\Rightarrow$ ${\sin ^2}\theta = \frac{1}{2} - \frac{1}{{\sqrt 5 }}$

$\Rightarrow$ $\sin \theta = \sqrt {\frac{1}{2} - \frac{1}{{\sqrt 5 }}}$

$therefore, {\cos ^2}\theta = 1 - {\sin ^2}\theta$
$= 1 - \frac{1}{2} + \frac{1}{{\sqrt 5 }} = \frac{1}{2} + \frac{1}{{\sqrt 5 }}$

$\Rightarrow$ $\cos \theta = \sqrt {\frac{1}{2} + \frac{1}{{\sqrt 5 }}}$

$therefore, \tan \theta = \sqrt {\frac{{\frac{1}{2} - \frac{1}{{\sqrt 5 }}}}{{\frac{1}{2} + \frac{1}{{\sqrt 5 }}}}} = \sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}}$

$\Rightarrow$ $\theta = {\tan ^{ - 1}}\sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}} = \frac{1}{2}{\cos ^{ - 1}}\frac{2}{{\sqrt 5 }}$

$therefore, \tan \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{2}{{\sqrt 5 }}} \right) = \tan {\tan ^{ - 1}}\sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}}}$

$= \sqrt {\frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \cdot \frac{{\sqrt 5 - 2}}{{\sqrt 5 - 2}}}$

$= \sqrt {\frac{{{{(\sqrt 5 - 2)}^2}}}{{5 - 4}}} = \sqrt 5 - 2$