# How do you find the exact value of  arctan(tan ((-7pi)/6))?

$\tan \left(\theta + n \pi\right) = \tan \left(\theta\right)$ for any $n \in \mathbb{Z}$
So to make the definition of $\arctan$ unique, its range is defined to be $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$
$\frac{- 7 \pi}{6} + \pi = - \frac{\pi}{6}$ lies in the range of $\arctan$ and is of the form $\frac{- 7 \pi}{6} + n \pi$
So $\arctan \left(\tan \left(\frac{- 7 \pi}{6}\right)\right) = - \frac{\pi}{6}$