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