${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x},\;x \ne 0$

Putting $x = \tan \theta ,$ we get
${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right) = {\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right]$

$= {\tan ^{ - 1}}\left( {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

$= {\tan ^{ - 1}}\left[ {\frac{{2{{\sin }^2}(\theta /2)}}{{2\sin (\theta /2)\cos (\theta /2)}}} \right] = {\tan ^{ - 1}}\left( {\tan \left( {\frac{\theta }{2}} \right)} \right) = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$

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