Question

Simplify
${\tan }^{-1}$ (tan $\frac{8\pi }{7}$).

Answer 1

We know that range of $ta{n}^{-1}x$ is $(-\frac{\pi }{2},\frac{\pi }{2})$

$\tan \left(\frac{8\pi }{7}\right)$

$=\tan (\pi +\frac{\pi }{7})$

$=\tan \left(\frac{\pi }{7}\right)$

Hence-

${\tan }^{-1}\left(\tan \left(\frac{8\pi }{7}\right)\right)$

$={\tan }^{-1}\left(\tan \left(\frac{\pi }{7}\right)\right)$

$=\frac{\pi }{7}$