Sum of the reciprocal of sine squared

I encountered an interesting identity when doing physics homework, that is,
$$ \sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } = \frac{N^2-1}{3}. $$

How is this identity derived? Are there any more related identities?

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From my answer here: Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$

Convert this to the problem of finding

$$\sum_{n=1}^N \frac{1}{1 – \cos \frac{2\pi n}{N}}$$

Convert this to the problem of getting a Chebyshev polynomial of which the roots are $\cos \frac{2\pi n}{N}$ (perhaps after using $\cos \frac{2 \pi (N-n)}{N} = \cos \frac{2 \pi n}{N}$ and getting a polynomial for a subset)

And using the fact the for any polynomial $P(x)$ with roots $r_i$ we have that

$$ \sum_{j=1}^{n} \frac{1}{x – r_j} = \frac{P'(x)}{P(x)}$$

Another (very similar approach) would be to start with

$$\frac{2}{\sin^2 x} = \frac{1}{1+\cos x} + \frac{1}{1-\cos x}$$
Note: I haven’t worked out the details, but I am pretty sure this would work.

Let’s denote $ S_N=\sum_{n=1}^{N-1} \frac{1}{\sin^2 \dfrac{\pi n}{N} } $

Consider an equality:

$$\frac{\pi^2}{\sin^2(\pi s)}=\int_0^{\infty}\frac{x^{s-1}}{1-x}\ln\frac{1}{x}dx;0<s<1$$

Because of $0<\frac{n}{N}<1$, the integral form applies. Thus:

$$S_N=\frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\sum_{n=1}^{N-1}x^{\frac{n}{N}-1}dx=$$

$$= \frac{1}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{x^{\frac{1}{N}-1}-1}{1-x^{\frac{1}{N}}}dx=$$

$$=\frac{N^2}{{\pi}^2} \int_0^{\infty}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx=$$

$$=2\frac{N^2}{{\pi}^2} \int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{3} $$

because of $$\int_0^{1}\frac{\ln\frac{1}{x}}{1-x}\frac{1-x^{N-1}}{1-x^{N}}dx= \frac{N^2-1}{N^2}\frac{{\pi}^2}{6}$$