## Precalculus (6th Edition)

$y= -\displaystyle \frac{\pi}{3}$
By definition of arctan $(\tan^{-1})$, y is the number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ such that $\tan y=-\sqrt{3}.$ In quadrant I, $\displaystyle \tan(\frac{\pi}{3})=\sqrt{3}$, so in quadrant IV, $\displaystyle \tan(-\frac{\pi}{3})=-\sqrt{3}.$ $-\displaystyle \frac{\pi}{3}\in (-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ so $y= -\displaystyle \frac{\pi}{3}$