# How do you find the exact value of arc tan(-sqrt3)?

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
Since $\sin \left(- \theta\right) = - \sin \theta$ and $\cos \left(- \theta\right) = \cos \theta$ for any $\theta$, we have
$\tan \left(- \frac{\pi}{3}\right) = \sin \frac{- \frac{\pi}{3}}{\cos} \left(- \frac{\pi}{3}\right) = \frac{- \sin \left(\frac{\pi}{3}\right)}{\cos} \left(\frac{\pi}{3}\right)$
$= - \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = - \frac{\sqrt{3}}{1} = - \sqrt{3}$
So $\arctan \left(- \sqrt{3}\right) = - \frac{\pi}{3}$