# How do you evaluate tan^-1(sqrt3/3)?

Aug 27, 2016

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

#### Explanation:

The Defn. of ${\tan}^{-} 1$ function is :

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

As $\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} , \mathmr{and} , \frac{\pi}{6} \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, we get,

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