Answer
See below.
Work Step by Step
1. To show that the period of $f(\theta)=csc\theta=\frac{1}{sin\theta}$ is $2\pi$, since we know $2\pi$ is a period, we only need to show that there is no other period less than $2\pi$.
2. Assume there is another period $p$ with $0\lt p\lt 2\pi$ and $csc(\theta+p)=csc\theta$ for all $\theta$, let $\theta=\frac{\pi}{2}$, we have $csc(p+\frac{\pi}{2})=csc(\frac{\pi}{2})=1$ or $sin(p+\frac{\pi}{2})=1$ thus $cos(p)=1$.
3. As $0\lt p\lt 2\pi$, there is no such value $p$ satisfies the above result.
4. Since the results of steps 2 and 3 are in contradiction to each other, thus such $p$ value does not exist, which proves that the minimum period is $2\pi$.
