Answer
$y= -\displaystyle \frac{\pi}{3}$
Work Step by Step
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 $y= -\displaystyle \frac{\pi}{3}$
