## Algebra 2 (1st Edition)

$\theta=-\dfrac{\pi}{6}$ or, $-30^{\circ}$
We are given $\tan^{-1} (-\dfrac{\sqrt 3}{3})$ We know that the trigonometric function $\tan^{-1}$ has domain R and the range is: $[-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$. So, $\theta = \tan^{-1}(-\dfrac{\sqrt 3}{3})$ This implies that: $\theta=-\dfrac{\pi}{6}$ or, $-30^{\circ}$