# How do you evaluate tan^-1(tan((13pi)/6))?

Jan 9, 2017

$\frac{\pi}{6}$.

#### Explanation:

${\tan}^{-} 1 \left\{\tan \left(\frac{13 \pi}{6}\right)\right\} = {\tan}^{-} 1 \left\{\tan \left(\frac{12 \pi + \pi}{6}\right)\right\}$

$= {\tan}^{-} 1 \left\{\tan \left(2 \pi + \frac{\pi}{6}\right)\right\}$

$= {\tan}^{-} 1 \left\{\tan \left(\frac{\pi}{6}\right)\right\} \ldots \left[\because , \tan \left(2 \pi + x\right) = \tan x\right]$

$= \frac{\pi}{6}$

In the last step, we have used the following Defn. of ${\tan}^{-} 1$ fun.:

$\theta = {\tan}^{-} 1 x , x \in \mathbb{R} \iff \tan \theta = x , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$