# How do you evaluate Tan[arccos ( - sqrt ( 3 ) / 2 ) ]?

Nov 14, 2015

$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{\sqrt{3}}{3}$

#### Explanation:

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$
$\sin \left(\theta\right) = \pm \sqrt{1 - {\cos}^{2} \left(\theta\right)}$

During the arccosine range, the sine is always positive so

$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = \frac{\sqrt{1 - {\left(- \frac{\sqrt{3}}{2}\right)}^{2}}}{- \frac{\sqrt{3}}{2}}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{2 \cdot \sqrt{1 - \frac{3}{4}}}{\sqrt{3}}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{2 \cdot \sqrt{3} \cdot \sqrt{1 - \frac{3}{4}}}{3}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{2 \cdot \sqrt{3} \cdot \sqrt{\frac{4 - 3}{4}}}{3}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{2 \cdot \sqrt{3} \cdot \sqrt{\frac{1}{4}}}{3}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{2 \cdot \sqrt{3} \cdot \left(\frac{1}{2}\right)}{3}$
$\tan \left(\arccos \left(- \frac{\sqrt{3}}{2}\right)\right) = - \frac{\sqrt{3}}{3}$