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.

Using the partial fractions identity, it's possible to split the summation into two
$$ \frac{2}{\sin^2 \theta} = \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta}$$
Now our sum over the nth roots
$$ (\star) =\sum_{k=1}^{m-1} \frac{1}{\sin^2 \frac{\pi k }{n}}
= \frac{1}{2}\sum_{k=1}^{m-1} \frac{1}{1-\cos \frac{\pi k }{n}}
+ \frac{1}{2}\sum_{k=1}^{m-1} \frac{1}{1+\cos \frac{\pi k }{n}}
$$
If we can find the right polynomial $P(x)$, we can find our sum using the identity for the logarithmic derivative. We need to plug in $x=1$ and $x=-1$ and subtract.
$$ \frac{P'(x)}{P(x)} =\sum \frac{1}{x-r_i} \text{ then }
\sum \frac{1}{1-r_i} +  \sum \frac{1}{1+r_i} 
= \frac{P'(1)}{P(1)} - \frac{P'(-1)}{P(-1)} $$
My first guess is the Chebyshev polynomials satisfy $T_n(\cos \theta)  = \cos n \theta $.
The roots of $T_m(x) = 0$ are $x = \cos \frac{2\pi k}{m}$ with $k = 0,1,2, \dots, m-1$.
Instead, we need a polynomial with roots of $x = \cos \frac{\pi k}{m}$ which is the derivative $\boxed{P(x)=T'_m(x)}$ (Chebyshev polynomial of second kind)

In the case of the Chebyshev polynomial I got these two expressions for the values of the derivative at $x=1$:
Let $T(\cos \theta) = \cos n \theta$.  Then set $\theta = 0$:
$$ T(1) = 1 $$
If we take the derivative (whose zeros have roots exactly where we want)
$$\boxed{ \sin \theta \; T'(\cos \theta) = -n \sin n \theta} \hspace{0.25in}\text{ so that }\hspace{0.25in}
 \boxed{\displaystyle T'(\cos \theta) = -\frac{n \sin n \theta}{\sin \theta} \bigg|_{\theta=0}= -n^2}$$
and for the second derivative, try the quotient rule:
\begin{eqnarray} T''(\cos \theta) &=& n \cdot \frac{n \cos n \theta \sin \theta - \sin n \theta \cos \theta}{\sin^3 \theta}  \\
&=\bigg|_{\theta \approx 0}& n \cdot \frac{n (1 - n^2 \theta^2 /2)(\theta - \theta^3 /6) - (n \theta - (n\theta)^3/6)(1 - \theta^2 /2)}{\theta^3} \\ &=\bigg|_{\theta \approx 0}& \frac{n^2 - n^4 }{3}\end{eqnarray} 
If we set $\theta = \pi$, just get a $(-1)^n$ factor.  In general your sum is 
$$ (\star) =  \frac{T'(1)}{T''(1)} = \frac{T'(-1)}{T''(-1)} =
\frac{1}{2}\left( \frac{T'(1)}{T''(1)} + \frac{T'(-1)}{T''(-1)}\right) =  \frac{\tfrac{1}{3}(n^4 - n^2)}{n^2} = \frac{n^2 - 1 }{3}$$

COMMENTS It is surprisingly hard to pin all the details down.  See also: Sum of the reciprocal of sine squared
