Prove that for every natural number $n$ and for every real numbers Prove that for every natural number $n$ and for every real numbers $x\neq \dfrac {k\pi}{2^t }$ $(t=0,1,....n;)$ (where $k$ is any integer)
$$\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx$$

Since 
\begin{align}
\frac{1}{\sin 2x}=\frac{2{\cos}^2x-\cos2x}{2\sin x\cos x}=\cot x-\cot 2x,
\end{align}
then, if we regard $x$ as $2x$
\begin{align}
\frac{1}{\sin 4x}=\frac{2{\cos}^22x-\cos 4x}{2\sin 2x \cos 2x}=\cot 2x-\cot 4x.
\end{align}
Similarly, repeat the above step until $n$,
we have 
\begin{align}
LHS=\cot x-\cot 2x+\cot 2x-\cot 4x+...+\cot 2^{n-1}x-\cot 2^nx=\cot x-\cot 2^nx=RHS.
\end{align}
