## Calculus: Early Transcendentals 8th Edition

a) $\tan^{-1}\sqrt3=\frac{\pi}{3}$ b) $\arctan(-1)=-\frac{\pi}{4}$
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 $k=0, \theta=-\frac{\pi}{4}$ $\therefore \arctan(-1)=-\frac{\pi}{4}$