Find the value of ${\tan ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right) + {\cot ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) + {\tan ^{ - 1}}\left[ {\sin \left( {\frac{{ - \pi }}{2}} \right)} \right]$.

We have, ${\tan ^{ - 1}}\left( { - \frac{1}{{\sqrt 3 }}} \right) + {\cot ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right) + {\tan ^{ - 1}}\left[ {\sin \left( {\frac{{ - \pi }}{2}} \right)} \right]$

$= {\tan ^{ - 1}}\left( {\tan \frac{{5\pi }}{6}} \right) + {\cot ^{ - 1}}\left( {\cot \frac{\pi }{3}} \right) + {\tan ^{ - 1}}( - 1)$

$= {\tan ^{ - 1}}\left[ {\tan \left( {\pi - \frac{\pi }{6}} \right)} \right] + {\cot ^{ - 1}}\left[ {\cot \left( {\frac{\pi }{3}} \right)} \right] + {\tan ^{ - 1}}\left[ {\tan \left( {\pi - \frac{\pi }{4}} \right)} \right]$

$= {\tan ^{ - 1}}\left( { - \tan \frac{\pi }{6}} \right) + {\cot ^{ - 1}}\left( {\cot \frac{\pi }{3}} \right) + {\tan ^{ - 1}}\left( { - \tan \frac{\pi }{4}} \right)$

$= - \frac{\pi }{6} + \frac{\pi }{3} - \frac{\pi }{4} = \frac{{ - 2\pi + 4\pi - 3\pi }}{{12}}$
$= \frac{{ - 5\pi + 4\pi }}{{12}} = - \frac{\pi }{{12}}$