Answer
$a.\displaystyle \quad \frac{3\pi}{4}$
$b.\displaystyle \quad \frac{\pi}{6}$
$c.\displaystyle \quad \frac{2\pi}{3}$
Work Step by Step
$y=\cot^{-1}x$ is the number in $(0, \pi)$ for which $\cot y=x.$
Use reference angles in quadrant I, with $\cot(\pi-t)=-\cot t$
$(a)$
In Q I,
$\displaystyle \cot(\frac{\pi}{4})=1$, so
$\displaystyle \cot(\pi-\frac{\pi}{4})=-1 \quad\Rightarrow\quad \cot^{-1}(-1)=\frac{3\pi}{4}$
$(b)$
In Q I,
$\displaystyle \cot(\frac{\pi}{6})=\sqrt{3}$, so$ \displaystyle \quad \cot^{-1}(\sqrt{3})=\frac{\pi}{6}$
$(c)$
In Q I,
$\displaystyle \cot(\frac{\pi}{3})=\frac{1}{\sqrt{3}}$, so
$\displaystyle \cot(\pi-\frac{\pi}{3})=-\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \cot^{-1}(-\frac{1}{\sqrt{3}})=\frac{2\pi}{3}$
