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.

Setting $\tan mx=0$  in  Sum of tangent functions where arguments are in specific arithmetic series, 
we get $$\binom m1\tan x-\binom m3\tan^3x+\cdots=0\ \ \ \ (1)$$
Multiplying both sides by $\cot^mx,$  
$$\binom m1\cot^{m-1}x-\binom m3\cot^{m-3}x+\cdots=0\ \ \ \ (2)$$
$\tan mx=0\implies mx=n\pi$ where $n$ is any integer
$\implies$ the roots of $(2)$ are $\cot x$ where $x=\dfrac{n\pi}m$ where $1\le n\le m-1$
Using Vieta's formula, $$\sum_{r=1}^{m-1}\cot^2\dfrac{r\pi}m=\left(\sum_{r=1}^{m-1}\cot\dfrac{r\pi}m\right)^2-2\left(\sum_{r_1,r_2=1,r_1>r_2}^{m-1}\cot\dfrac{r_1\pi}m\cot\dfrac{r_2\pi}m\right)$$
$$=0-2\cdot\left(-\frac{\binom m3}{\binom m1}\right)=\cdots$$
$$\sum_{r=1}^{m-1}\frac1{\sin^2\dfrac{r\pi}m}=\sum_{r=1}^{m-1}\left(1+\cot^2\dfrac{r\pi}m\right)=m-1+\sum_{r=1}^{m-1}\cot^2\dfrac{r\pi}m=\cdots$$
Hope the steps are understandable
