Find the value: ^ -1 ( 7 6 ) - Vidyadip

Question

Find the value:
$\tan^{-1}\bigg(\tan\frac{7\pi}{6}\bigg)$

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Answer

We know that $\tan^{-1}(\tan x)=x$ if $x\in\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg),$ which is the principal value branch of $\tan^{-1}x.$
Here, $\frac{7\pi}{6}\notin\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg).$
Now, $\tan^{-1}\bigg(\tan\frac{7\pi}{6}\bigg)$ can be written as:
$\tan^{-1}\bigg(\tan\frac{7\pi}{6}\bigg)=\tan^{-1}\bigg[\tan\bigg(2\pi-\frac{5\pi}{6}\bigg)\bigg]$ $\left[\tan\left(2\pi-x\right)=-\tan x\right]$
$=\tan^{-1}\bigg[-\tan\bigg(\frac{5\pi}{6}\bigg)\bigg]=\tan^{-1}\bigg[\tan\bigg(-\frac{5\pi}{6}\bigg)\bigg]$
$=\tan^{-1}\bigg[\tan\bigg(\pi-\frac{5\pi}{6}\bigg)\bigg]$
$=\tan^{-1}\bigg[\tan\bigg(\frac{\pi}{6}\bigg)\bigg], \text{where}\frac{\pi}{6}\in\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)$
$\therefore\tan^{-1}\bigg(\tan\frac{7\pi}{6}\bigg)=\tan^{-1}\bigg(\tan\frac{\pi}{6}\bigg)=\frac{\pi}{6}$

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