Find the value of
$ \displaystyle {\tan ^{ – 1}}\left( {\tan \frac{5\pi}{6}} \right) + {\cos ^{ – 1}}\left( {\cos \frac{13\pi}{6}} \right) $
[NCERT, Ex.2.3, Q.1, Page.35]
Solution
We know that: $ \displaystyle \tan^{-1}(\tan x) = x,\quad x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) $ and $ \displaystyle \cos^{-1}(\cos x) = x,\quad x \in [0,\pi] $
$\therefore\ {\tan ^{ – 1}}\left( {\tan \frac{5\pi}{6}} \right) + {\cos ^{ – 1}}\left( {\cos \frac{13\pi}{6}} \right)$
$= {\tan ^{ – 1}}\left[ {\tan \left( \pi – \frac{\pi}{6} \right)} \right] + {\cos ^{ – 1}}\left[ {\cos \left( \pi + \frac{7\pi}{6} \right)} \right]$
$= {\tan ^{ – 1}}\left( -\tan \frac{\pi}{6} \right) + {\cos ^{ – 1}}\left( -\cos \frac{7\pi}{6} \right)$
$= -{\tan ^{ – 1}}\left( {\tan \frac{\pi}{6}} \right) + \pi – \left[ {\cos ^{ – 1}}\left( {\cos \frac{7\pi}{6}} \right) \right]$
Using $\cos^{-1}(-x) = \pi – \cos^{-1}(x),\ x \in [-1,1]$
$= -{\tan ^{ – 1}}\left( {\tan \frac{\pi}{6}} \right) + \pi – {\cos ^{ – 1}}\left[ {\cos \left( \pi + \frac{\pi}{6} \right)} \right]$
$= -{\tan ^{ – 1}}\left( {\tan \frac{\pi}{6}} \right) + \pi – \pi + {\cos ^{ – 1}}\left( {\cos \frac{\pi}{6}} \right)$
$= -\frac{\pi}{6} + 0 + \frac{\pi}{6} = 0$
Keywords: ${\tan ^{ – 1}}\left( {\tan \frac{5\pi}{6}} \right) + {\cos ^{ – 1}}\left( {\cos \frac{13\pi}{6}} \right)$