Answer
$a.\qquad-2 \\$
$b. \displaystyle \qquad \frac{2\sqrt{3}}{3} \\$
$c. \qquad \sqrt{3}$
Work Step by Step
See p.412: Evaluating trig. functions for any number:
1. find the reference number,
2. Find the sign , use table on p.412,
3. Find the value (using the reference number and sign)
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$a.\displaystyle \qquad \frac{7\pi}{6}=\pi+\frac{\pi}{6}$ lies in quadrant III,
reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \csc\frac{\pi}{6}=2$ (table, p.410)
In quadrant III, csc is negative (see table on p.412), so
$\displaystyle \csc\frac{7\pi}{6}=-2.$
$b. \displaystyle \qquad -\frac{\pi}{6}=0-\frac{\pi}{6}$ lies in quadrant IV,
reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \sec\frac{\pi}{6}=\frac{2\sqrt{3}}{3}.$ (table, p.410)
In quadrant IV, sec is positive (see table on p.412), so
$\displaystyle \sec(-\frac{\pi}{6})=\frac{2\sqrt{3}}{3}$
$c. \displaystyle \qquad -\frac{\pi}{6}=-\pi+\frac{\pi}{6}$ lies in quadrant III,
reference number : $\displaystyle \frac{\pi}{6}$, with $\displaystyle \cot\frac{\pi}{6}=\sqrt{3}.$ (table, p.410).
In quadrant III, $\cot t$ is positive (see table on p.412), so
$\displaystyle \cot(-\frac{\pi}{6})=\sqrt{3}$