Answer
$-\frac{\sqrt 3}{2},\frac{1}{2},-\sqrt 3,-\frac{\sqrt {3}}{3},2,-\frac{2\sqrt 3}{3}$
Work Step by Step
1. Given $\theta=1020^\circ=3\times360^\circ-60^\circ$, we can identify it is in quadrant IV with a reference angle $60^\circ$, we have:
2. $sin\theta=-sin60^\circ=-\frac{\sqrt 3}{2}$
3. $cos\theta=cos60^\circ=\frac{1}{2}$
4. $tan\theta=-tan60^\circ=-\sqrt 3$
5. $cot\theta=\frac{1}{tan\theta}=-\frac{\sqrt {3}}{3}$
6. $sec\theta=\frac{1}{cos\theta}=2$
7. $csc\theta=\frac{1}{sin\theta}=-\frac{2\sqrt 3}{3}$
