Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$ 
Question : Is the following true for any $m\in\mathbb N$?
  $$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$

Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let 
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies 
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.

(First time I write in a math blog, so forgive me if my contribution ends up being useless)
I recently bumped into this same identity while working on Fourier transforms. By these means you can show in fact that
$$\sum_{k=-\infty}^{+\infty}\frac{1}{(x-k)^2}=\frac{\pi^2}{\sin^2 (\pi x)}.\tag{*}\label{*}$$
Letting $x=\frac13$ yields
$$\begin{align}\sum_{k=-\infty}^{+\infty}\frac{1}{(3k-1)^2}&= \sum_{k=0}^{+\infty}\left[\frac{1}{(3k+1)^2}+ \frac{1}{(3k+2)^2} \right]=\\
&=\sum_{k=1}^{+\infty} \frac{1}{k^2} - \sum_{k=1}^{+\infty}\frac{1}{(3k)^2}=\\
&=\frac{8}{9}\sum_{k=1}^{+\infty} \frac{1}{k^2}\\
&\stackrel{\eqref{*}}= \frac{4\pi^2}{27}\end{align},$$
which in the end gives
$$\sum_{k=1}^{+\infty} \frac{1}{k^2} = \frac{\pi^2}{6}.\tag{**}\label{**}$$
Now, back to \eqref{*}, we can generalize and take $x=m/n$, with $n$ and $m$ positive integers, obtaining
$$\sum_{k=0}^{+\infty}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]=\frac{\pi^2}{n^2 \sin^2\left(\frac{\pi m}{n}\right)}.$$
Finally taking the sum of LHS and RHS of last equation for $m=1,2,\dots,n-1$ yields
$$\begin{align}\sum_{m=1}^{n-1}\sum_{k=0}^{+\infty}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]&=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\
\sum_{k=0}^{+\infty}\sum_{m=1}^{n-1}\left[\frac{1}{(nk+m)^2} + \frac{1}{(nk+n-m)^2}\right]&=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\
2\sum_{k=1}^{+\infty}\frac{1}{k^2} - 2\sum_{k=1}^{+\infty}\frac{1}{(nk)^2} &=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}\\
\frac{2(n^2-1)}{n^2}\sum_{k=1}^{+\infty}\frac{1}{k^2} &=\frac{\pi^2}{n^2} \sum_{m=1}^{n-1}\frac{1}{\sin^2\left(\frac{\pi m}{n}\right)}.
\end{align}$$
Using then \eqref{**} leads to the result
$$\sum_{m=1}^{n-1} \frac{1}{\sin^2\left(\frac{\pi m}{n}\right)} = \frac{n^2-1}{3}.$$
