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
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*}
