Question 4
(YUG) Prove the following equality:
$$\frac1{\sin2x}+\frac1{\sin4x}+\frac1{\sin8x}+\cdots+\frac1{\sin2^nx}=\cot x-\cot2^nx,$$
where $n\in\mathbb N$ and $x\notin\pi\mathbb Z/2^k$ for every $k\in\mathbb N$.
$$\frac1{\sin2x}+\frac1{\sin4x}+\frac1{\sin8x}+\cdots+\frac1{\sin2^nx}=\cot x-\cot2^nx,$$
where $n\in\mathbb N$ and $x\notin\pi\mathbb Z/2^k$ for every $k\in\mathbb N$.