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)
$$\dfrac {1}{\sin 2x}+\dfrac {1}{\sin 4x}+.....+\dfrac {1}{\sin 2^n x}=\cot x-\cot 2^nx$$

Here's a proof by induction 
This is true for base case $n=1$
Assume true for $n$ ...So $$\cot x-\cot2^{n+1}x=\cot x-\cot(2(2^{n}x))=\cot x-\frac12((\cot(2^nx)-\tan(2^nx))$$$$= \cot x-\frac12\cot(2^nx)+\frac12\tan(2^nx)=\cot x-\cot(2^nx)+\frac12(\cot(2^nx)+\tan(2^nx))$$
Note that
$$\frac12(\cot(2^nx)+\tan(2^nx))=\frac{1}{\sin(2^{n+1}x)}$$
where I used $\cot x +\tan x=\frac{1}{\sin x\cos x}$ and $\cot2x=\frac12(\cot x-\tan x)$
So..can you complete now?
