Answer
See answers below.
Work Step by Step
1. $240^\circ$ is in quadrant III with a reference angle $\theta=240^\circ-180^\circ=60^\circ$
2. We have $sin240^\circ=-sin\theta=-sin60^\circ=-\frac{\sqrt 3}{2}$
3. We have $cos240^\circ=-cos\theta=-cos60^\circ=-\frac{1}{2}$
4. We have $tan240^\circ=tan\theta=tan60^\circ=\sqrt 3$
5. We have $cot240^\circ=cot\theta=cot60^\circ=\frac{\sqrt 3}{3}$
6. We have $sec240^\circ=-sec\theta=-\frac{1}{cos\theta}=-2$
7. We have $csc240^\circ=-csc\theta=-\frac{1}{sin\theta}=-\frac{2\sqrt 3}{3}$
