Answer
a) $\cot^{-1}(-\sqrt3)=\frac{5\pi}{6}$
b) $\sec^{-1}2=\frac{\pi}{3}$
Work Step by Step
a) let $\cot^{-1}(-\sqrt3)=\theta, 0\lt\theta\lt\pi$
$\cot\theta=-\sqrt3$
$\tan\theta=\frac{-1}{\sqrt3}$
$\theta=-\frac{\pi}{6}+k\pi$, where k is an integer.
$\because 0\lt\theta\lt\pi,$ let $k=1, \theta=\frac{5\pi}{6}$
$\therefore \cot^{-1}(-\sqrt3)=\frac{5\pi}{6}$
b) let $\sec^{-1}2=\theta, 0\leq\theta\leq\pi, \theta\neq\frac{\pi}{2}$
$\sec\theta=2$
$\cos\theta=\frac{1}{2}$
$\theta=\pm\frac{\pi}{3}+2k\pi$, where k is an integer.
$\because 0\leq\theta\leq\pi, \theta\neq\frac{\pi}{2},$ let $k=0, \theta=\frac{\pi}{3}$
$\therefore \sec^{-1}2=\frac{\pi}{3}$
