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

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#### Explanation

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#### Explanation:

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mason m Share
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 $\tan \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$.

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

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

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