Answer
$a.\qquad-\sqrt{2} \\$
$b. \displaystyle \qquad -\frac{1}{2} \\$
$c. \displaystyle \qquad -\frac{\sqrt{3}}{3}$
Work Step by Step
See p.412: Evaluating trig. functions for any number:
1. find the reference number,
2. Find the sign
3. Find the value (using the reference number and sign)
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$a.\displaystyle \qquad \frac{3\pi}{4}=\pi-\frac{\pi}{4}$ lies in quadrant II,
reference number : $\displaystyle \frac{\pi}{4}$, with $\displaystyle \sec\frac{\pi}{4}=\sqrt{2} $ (table, p.410)
In quadrant II, $\sec t$ is negative , (table, p.412) so
$\displaystyle \sec\frac{3\pi}{4}=-\sqrt{2}.$
b$.\displaystyle \qquad -\frac{2\pi}{3}=-\pi+\frac{\pi}{3}$ lies in quadrant III,
reference number : $\displaystyle \frac{\pi}{3}$, with $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2} $ (table, p.410)
In quadrant II, $\cos t$ is negative , (table, p.412) so
$\displaystyle \cos(-\frac{2\pi}{3})=-\frac{1}{2}.$
$ c.\qquad -\displaystyle \frac{7\pi}{6}=-\pi-\frac{\pi}{6}$ lies in quadrant II,
reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}.$ (table, p.410)
In quadrant II, $\tan t$ is negative (table, p.412), so
$\displaystyle \tan(-\frac{7\pi}{6})=-\frac{\sqrt{3}}{3}.$