${\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }},\;|x|\; > 1$

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

$= {\tan ^{ - 1}}\left( {\frac{1}{{\tan \theta }}} \right) = {\tan ^{ - 1}}(\cot \theta ) = {\tan ^{ - 1}}\left\{ {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right\}$

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