Find sum of $\frac{1}{\sin\theta\cdot \sin2\theta} + \frac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$

Question

$$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$$ up to $n$ terms.

I tried but in vain

Answer 1

Note $$\dfrac{\sin{\theta}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\dfrac{\sin{((k+1)\theta-k\theta)}}{\sin{(k+1)\theta}\sin{(k\theta)}}=\cot{(k\theta)}-\cot{(k+1)\theta}$$

Answer 2

Here is a hint: the sum telescopes, with some careful manipulation:

$$\sin \theta = \sin((k+1)\theta - k\theta) = \sin(k+1)\theta \cos k\theta - \sin k\theta \cos (k+1)\theta.$$