Trigonometric series Cosecant^4 How to prove the following identity

$$
\text{cosec}^4{\frac{π}{n}}+\text{cosec}^4{\frac{2π}{n}}+\text{cosec}^4{\frac{3π}{n}} + \ldots+ \text{cosec}^4{\frac{(n-1)π}{n}} = \frac{(n^2+11)(n^2-1)}{45}
$$

I tried to simplify the sum but I finally get stuck with the same question I don't know how to start the problem
 A: The best way is probably appealing to Chebyshev polynomials and Vieta's formula.
Note that $x=\sin(k\pi/n)$, $k=0,\dots,2n-1$, are the roots to $\cos(n\arccos(1-2x^2))=1$, i.e., $T_n(1-2x^2)=1$, where $T_n$ is the Chebyshev polynomial of the first kind.
Now using
$$
T_n(y)=n\sum_{k=0}^n(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(1-y)^k
$$
we get
$$
0=T_n(1-2x^2)-1=-1+n\sum_{k=0}^n(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(2x^2)^k
$$
Hence
$$
0=-2n^2x^2+\frac23n^2(n^2-1)x^4-\frac4{45}n^2(n^4-5n^2+4)x^6+\dots
$$
Excluding the two zeros $k=0,n$ (which gives $x=0$), we have
$$
0=-2n^2+\frac23n^2(n^2-1)x^2-\frac4{45}n^2(n^4-5n^2+4)x^4+\dots
$$
So Vieta's formula gives
\begin{align*}
\sum_{k=1}^{n-1}\frac{1}{\sin^2(k\pi/n)}&=\frac13(n^2-1),\\
\sum_{1\leq k<l<n}\frac1{\sin^2(k\pi/n)\sin^2(l\pi/n)}&=\frac2{45}(n^4-5n^2+4)
\end{align*}
Hence
\begin{align*}
\sum_k\operatorname{cosec}^4(k\pi/n)&=\sum_k\frac{1}{\sin^4(k\pi/n)}\\
&=\left(\sum_k\frac{1}{\sin^2(k\pi/n)}\right)^2-2\sum_{k<l}\frac1{\sin^2(k\pi/n)\sin^2(l\pi/n)}\\
&=\left(\frac13(n^2-1)\right)^2-\frac4{45}(n^4-5n^2+4)\\
&=\frac1{45}(n^2+11)(n^2-1).
\end{align*}
