Question

Solve:
${\tan }^{-1}\left(\tan \frac{7\pi }{6}\right)$

Answer 1

$\frac{7\pi }{6}\in (\pi ,\frac{3\pi }{2})$

So tan${}^{-1}\frac{7\pi }{6}=\frac{7\pi }{6}-\pi =\frac{\pi }{6}$

The reason we do so is that $ta{n}^{-1}m\in (\frac{-\pi }{2},\frac{\pi }{2})$ So we have to find the angle m in $(\frac{-\pi }{2},\frac{\pi }{2})$ for which $tanm=tan\frac{7\pi }{6}$