<s>[INST] Prove a Trigonometric Series Question:
$$\cot^{2}\frac{\pi }{2m+1}+\cot^{2}\frac{2\pi }{2m+1}+\cdots+\cot^{2}\frac{m\pi }{2m+1}=\frac{m(2m-1)}{3}$$
$m$ is a positive integer.
Attempt:
I started by showing that
$$\sin(2m+1)\theta =\binom{2m+1}{1}\cos^{2m}\theta \sin\theta -\binom{2m+1}{3}\cos^{2m-2}\theta \sin^{3}\theta +\cdots+(-1)^{m}\sin^{2m+1}\theta$$
By expanding $\left(\text{cis }\theta\right)^{2m+1}$ and then equating the imaginary part.
 [/INST] Divide this equality by $\sin^{2m+1}\theta$ to get
$$
\frac{\sin(2m+1)\theta}{\sin^{2m+1}\theta}=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1}\cot^{2k} \theta\tag{1}
$$
Denote
$$
P_m(x)=\sum\limits_{k=0}^m (-1)^k{2m+1 \choose 2k+1} x^m
$$
This is polynomial of degree $m$. From $(1)$ we see that $P_m$ have $m$ roots $x_k=\cot^2\frac{\pi k}{2m+1}$. By Vietas formula their sum equals to 
$$
-\frac{{2m+1 \choose 3}}{{2m+1 \choose 1}}=\frac{m(2m-1)}{3}
$$
So we get
$$
\sum\limits_{k=1}^m\cot^2\frac{\pi k}{2m+1}=\frac{(2m-1)m}{3}
$$
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