<s>[INST] Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $ 
Prove 
  $$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
  where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.

My try. If $n=2$, we have
$$\begin{align}
\frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt]
&= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x}
\end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.
 [/INST] Skipping over the base case ($n=1$) for the moment, let's do the inductive step:  If we take the equation
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^nx}=\cot x-\cot2^nx$$
and replace $x$ with $2x$, we get
$${1\over\sin4x}+{1\over\sin8x}+\cdots+{1\over\sin2^{n+1}x}=\cot2x-\cot2^{n+1}x$$
Adding $1/\sin2x$ to both sides now gives
$${1\over\sin2x}+{1\over\sin4x}+\cdots+{1\over\sin2^{n+1}x}={1\over\sin2x}+\cot2x-\cot2^{n+1}x$$
so, to complete the inductive step, it suffices to prove
$${1\over\sin2x}+\cot2x=\cot x$$
But this is the base case, just rewritten with the $\cot2x$ moved to the left hand side!  So let's prove it:
$$\begin{align}
{1\over\sin2x}+\cot2x&={1+\cos2x\over\sin2x}\\
&={1+(2\cos^2x-1)\over2\sin x\cos x}\\
&={\cos x\over\sin x}\\
&=\cot x
\end{align}$$
</s>