Answer
a) $\tan^{-1}\sqrt3=\frac{\pi}{3}$
b) $\arctan(-1)=-\frac{\pi}{4}$
Work Step by Step
a) let $\tan^{-1}\sqrt3=\theta , -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$
$\tan\theta=\sqrt3$
$\theta=\frac{\pi}{3}+k\pi$
$\because -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, let $k=0, \theta=\frac{\pi}{3}$
$\therefore \tan^{-1}\sqrt3=\frac{\pi}{3}$
b) let $\arctan(-1) =\theta, -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$
$\tan\theta=-1$
$\theta=-\frac{\pi}{4}+k\pi$
$\because -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, let $k=0, \theta=-\frac{\pi}{4}$
$\therefore \arctan(-1)=-\frac{\pi}{4}$
