Find the value of ${\tan ^{ – 1}}\left( {\tan \frac{{5\pi }}{6}} \right) + {\cos ^{ – 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$.
[NCERT,Exemplar.3,Q.1,Page.35]

We know that, ${\tan ^{ – 1}}\tan x = x;x \in \left( { – \frac{\pi }{2},\frac{\pi }{2}} \right)$ and ${\cos ^{ – 1}}\cos x = x;x \in [0,\pi ]$

$therefore, {\tan ^{ – 1}}\left( {\tan \frac{{5\pi }}{6}} \right) + {\cos ^{ – 1}}\left( {\cos \frac{{13\pi }}{6}} \right)$

$ = {\tan ^{ – 1}}\left[ {\tan \left( {\pi – \frac{\pi }{6}} \right)} \right] + {\cos ^{ – 1}}\left[ {\cos \left( {\pi + \frac{{7\pi }}{6}} \right)} \right]$

$ = {\tan ^{ – 1}}\left( { – \tan \frac{\pi }{6}} \right) + {\cos ^{ – 1}}\left( { – \cos \frac{{7\pi }}{6}} \right)$

$ = – {\tan ^{ – 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi – \left[ {{{\cos }^{ – 1}}\cos \left( {\frac{{7\pi }}{6}} \right)} \right]$

and $\left. {{{\cos }^{ – 1}}( – x) = \pi – {{\cos }^{ – 1}}x;x \in [ – 1,1]} \right\}$

$ = – {\tan ^{ – 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi – {\cos ^{ – 1}}\left[ {\cos \left( {\pi + \frac{\pi }{6}} \right)} \right]$

$ = – {\tan ^{ – 1}}\left( {\tan \frac{\pi }{6}} \right) + \pi – \pi + {\cos ^{ – 1}}\left( {\cos \frac{\pi }{6}} \right)$

$ = – \frac{\pi }{6} + 0 + \frac{\pi }{6} = 0$

Buy Mathematics E -Books : https://mathstudy.in/shop/