Chapter 9.4, Problem 9E

Solution

To find: the value of the given expression both in degrees and radians.

  $\theta =-\frac{\pi }{6}$ or $\theta =-{30}^{\circ }$ .

Given:

  ${\tan }^{-1}\left(-\frac{\sqrt{3}}{3}\right)$ .

Calculation:

When $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$ , or $-{90}^{\circ }<\theta <{90}^{\circ }$ , the angle whose tangent is $-\frac{\sqrt{3}}{3}$ is,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\\ =-\frac{\pi }{6}\end{array}$

Or,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\\ =-{30}^{\circ }\end{array}$