1969
Problem - 3284
Prove for every positive integer $n$ and real number $x\ne \frac{k\pi}{2^t}$ where $t =0, 1, 2,\cdots$ and $k$ is an integer, the following relation always holds:
$$\frac{1}{\sin 2x}+\frac{1}{\sin 4x} + \cdots +\frac{1}{\sin 2^nx}=\frac{1}{\tan x}-\frac{1}{\tan 2^nx}$$
Let $z=\cos x + i\sin x$. Then it is easy to verify that $$\tan x = -i\cdot\frac{z^2-1}{z^2+1}\implies \frac{1}{\tan x} = i\cdot\frac{z^2 + 1}{z^2-1}$$
It follows the claim is equivalent to
$$\frac{2\cdot z^2}{z^4-1}+\frac{2\cdot z^4}{z^8-1}+\cdots+\frac{2\cdot z^{2^n}}{z^{2^{n+1}}-1}=\frac{z^2+1}{z^2-1}-\frac{z^{2^{n+1}}-1}{z^{2^{n+1}}-1}$$
or
$$\frac{2\cdot z^2}{z^4-1}+\frac{2\cdot z^4}{z^8-1}+\cdots+\frac{2\cdot z^{2^n}}{z^{2^{n+1}}-1}+\frac{z^{2^{n+1}}-1}{z^{2^{n+1}}-1}=\frac{z^2+1}{z^2-1}$$
This above relation indeed holds if we evaluate it backwards by noticing
\begin{align}
&\frac{2\cdot z^{2^n}}{z^{2^{n+1}}-1}+\frac{z^{2^{n+1}}-1}{z^{2^{n+1}}-1}\\
= &\frac{z^{2^{n+1}}-1+2\cdot z^{2^n}}{z^{2^{n+1}}-1}\\
= &\frac{\Big(z^{2^n}+1\Big)^2}{\Big(z^{2^n}+1\Big)\Big(z^{2^n}-1\Big)}\\
= &\frac{z^{2^n}-1}{z^{2^n}-1}
\end{align}
Therefore the left side telescopes which will results in $\frac{z^2+1}{z^-1}$. This equals the right side of the above relation.