${\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)$

As we know that the range of principal value branch of ${\tan ^{ - 1}},\;\;{\cos ^{ - 1}}\;\;and\;\;{\sin ^{ - 1}}$ are
$\left( { - \frac{\pi }{2},\;\,\frac{\pi }{2}} \right),\;[0,\;\pi ]\;$ and $\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$ respectively.

Let ${\tan ^{ - 1}}(1) = x \Rightarrow 1 = \tan x$

Then, $1 = \tan \left( {\frac{\pi }{4}} \right),\;\,where\;\;\frac{\pi }{4} \in \left( { - \frac{\pi }{2},\;\frac{\pi }{2}} \right)$

Let ${\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) = y \Rightarrow - \frac{1}{2} = \cos y$

Then, $- \frac{1}{2} = - \cos \left( {\frac{\pi }{3}} \right) = \cos \left( {\pi - \frac{\pi }{3}} \right) = \cos \frac{{2\pi }}{3},$

Where $\frac{{2\pi }}{3} \in [0,\;\pi ]$

Let ${\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = z \Rightarrow - \frac{1}{2} = \sin z$

Then, $- \frac{1}{2} = \sin \left( { - \frac{\pi }{6}} \right),\;where\; - \frac{\pi }{6} \in \left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$

So, ${\tan ^{ - 1}}(1) + {\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = \left( {\frac{\pi }{4} + \frac{{2\pi }}{3} - \frac{\pi }{6}} \right) = \frac{{3\pi }}{4}.$