## Precalculus (6th Edition) Blitzer

The exact value of the provided expression is $-\frac{\pi }{6}$.
Let us assume $\theta ={{\tan }^{-1}}\left( -\frac{\sqrt{3}}{3} \right)$. Then, we have $\tan \theta =\left( -\frac{\sqrt{3}}{3} \right)$ We know that for the tan function, the interval for the angle is $\left( -\frac{\pi }{2},\frac{\pi }{2} \right)$. Therefore, the only angle that satisfies $\tan \theta =\left( -\frac{\sqrt{3}}{3} \right)$ is $\left( -\frac{\pi }{6} \right)$. Hence, $\theta =\left( -\frac{\pi }{6} \right)$ and the exact value of the given expression is ${{\tan }^{-1}}\left( -\frac{\sqrt{3}}{3} \right)=-\frac{\pi }{6}$