# Find the principal value of the following $tan^{-1} \bigg( -\frac{1}{\sqrt 3} \bigg)$

Toolbox:
• The range of the principal value of $\; tan^{-1}x$ is $\left [ \large-\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Ans : $$tan^{-1}tan \large\frac{-\pi}{6}=\large\frac{-\pi}{6}$$
Let $tan^{-1}-\frac{1}{\sqrt 3} = x \Rightarrow tan (x) =\frac{1}{ -\sqrt 3}$
We know that the range of the principal value of $\; tan^{-1}x$ is $\left [ \large-\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$
Therefore, $tan (x) =\frac{1}{ -\sqrt 3} = - tan \large\frac{\pi}{6} = tan \large\frac{-\pi}{6}$
$\Rightarrow x=\large\frac{-\pi}{6}$, where $x \;\epsilon\;$ $\left [ \large-\frac{\pi}{2}, \large\frac{\pi}{2} \right ]$

edited Mar 19, 2013