Answer
$-\frac{\sqrt {39}}{8}, -\frac{5}{8}, \frac{\sqrt {39}}{5}, \frac{5\sqrt {39}}{39}, -\frac{8}{5}, -\frac{8\sqrt {39}}{39}$
Work Step by Step
1. Given $cos\theta=-\frac{5}{8}$ and $\theta$ in quadrant III, form a right triangle with sides $\sqrt {8^2-5^2}, 5, 8$ or $\sqrt {39}, 5, 8$, we have:
2. $sin\theta=-\frac{\sqrt {39}}{8}$
3. $cos\theta=-\frac{5}{8}$
4. $tan\theta=\frac{\sqrt {39}}{5}$
5. $cot\theta=\frac{1}{tan\theta}=\frac{5\sqrt {39}}{39}$
6. $sec\theta=\frac{1}{cos\theta}=-\frac{8}{5}$
7. $csc\theta=\frac{1}{sin\theta}=-\frac{8\sqrt {39}}{39}$
