Browse Questions

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

$\begin{array}{1 1} \frac{\pi}{6} \\ \frac{7 \pi}{6} \\ \frac{-7 \pi}{6} \\ \frac{-\pi}{6} \end{array}$

Toolbox:
• Principal interval of tan is$(\large-\frac{\pi}{2},\large\frac{\pi}{2})$
• $tan(\pi+x)=tanx$
Given $tan^{-1} \bigg(tan \large\frac{7\pi}{6}\bigg)$
$\large\frac{7\pi}{6}$ is not in the principal interval of $tan \rightarrow$ reduce it within the principal interval $\rightarrow$ $\large\frac{7\pi}{6}=\pi+\large\frac{\pi}{6}$

The given expression can be written as $tan^{-1}tan \bigg( \pi+\large\frac{\pi}{6} \bigg)$
By taking $x=\large\frac{\pi}{6},$ tan$(\pi+\large\frac{\pi}{6})=tan\large\frac{\pi}{6}$
$\Rightarrow tan^{-1}tan \large\frac{\pi}{6}=\large\frac{\pi}{6}$

edited Mar 15, 2013