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
$y=\sec^{-1}x$ is the number in $[0, \pi/2$) $\cup(\pi/2, \pi$] for which $\sec y=x.$
$(\displaystyle \sec y=x \Leftrightarrow \cos y=\frac{1}{x})$
Use reference angles in the 1st quadrant,
keeping in mind that $\cos(\pi-t)=-\cos t$
$(\mathrm{a})$
$\displaystyle \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\quad\Rightarrow\quad\cos(\pi-\frac{\pi}{4})=-\frac{1}{\sqrt{2}}\qquad \Rightarrow\quad\sec(\frac{3\pi}{4})=-\sqrt{2}$
$\displaystyle \Rightarrow\quad\sec^{-1}(-\sqrt{2})=\frac{3\pi}{4}$
$(b)$
$\displaystyle \cos\frac{\pi}{6}=\frac{ \sqrt{3}}{2}\quad\Rightarrow\quad \sec(\frac{\pi}{6})=\frac{2}{\sqrt{3}}\quad\Rightarrow\quad\sec^{-1}(\frac{2}{\sqrt{3}})=\frac{\pi}{6}$
$(c)$
$\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}\quad\Rightarrow\quad \cos(\pi-\frac{\pi}{3})=\frac{1}{2}\quad\Rightarrow\quad\sec(\frac{2\pi}{3})=2$
$\displaystyle \Rightarrow\quad\sec(2)=\frac{2\pi}{3}$
