# How do you evaluate arctan(-1/sqrt3)?

Jul 11, 2016

$\arctan \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6.}$

#### Explanation:

Let $\arctan \left(- \frac{1}{\sqrt{3}}\right) = \theta , \theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

Then, by defn. of $\arctan$ fun, $\tan \theta = - \frac{1}{\sqrt{3}} < 0 ,$ so that $\theta \notin \left(0 , \frac{\pi}{2}\right)$

Now, $\tan \left(- \frac{\pi}{6}\right) = - \tan \left(\frac{\pi}{6}\right) = - \frac{1}{\sqrt{3}} ,$ where, $\left(- \frac{\pi}{6}\right) \in \left(- \frac{\pi}{2} , 0\right)$

Thus, $\tan \left(- \frac{\pi}{6}\right) = - \frac{1}{\sqrt{3}} = \tan \theta ,$ and, $\tan$ fun. is injective i.e., $1 - 1$ in $\left(- \frac{\pi}{2} , 0\right)$, we conclude that $\theta = \arctan \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6.}$