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

Nov 25, 2015

$\frac{\pi}{6}$

#### Explanation:

Another way of asking this question is: "the tangent of what angle gives me $\frac{\sqrt{3}}{3}$?"

We know that $\text{ } \tan \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$.

So then, $\text{ "" } {\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$

Note that even though tangent is positive in the third quadrant, and that the tangent of an infinite amount of coterminal angles is $\frac{\sqrt{3}}{3}$, we need to observe that the range of ${\tan}^{-} 1 \left(x\right)$ is $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, so only $\frac{\pi}{6}$ is a valid answer within this range.