Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi  (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreciated.

Like How to prove: $11=10^{12}+10^{7}-45\sum_{n=1}^{999}\csc^4\frac{n\pi}{1000}$
Question regarding $f(n)=\cot^2\left(\frac\pi n\right)+\cot^2\left(\frac{2\pi}n\right)+\cdots+\cot^2\left(\frac{(n-1)\pi}n\right)$
Observe that  $n\left(\frac{\pi  (2 k+1)}{2 n}+y\right)=ny+k\pi+\dfrac\pi2$
As $\tan nx=\dfrac{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}{\cot^nx-\binom n{n-2}\cot^{n-2}x+\binom n{n-4}\cot^{n-4}x-\cdots},$
if $\tan nx=\tan\left(ny+k\pi+\dfrac\pi2\right)=-\cot ny$
$$-\dfrac1{\tan ny}=\dfrac{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}{\cot^nx-\binom n{n-2}\cot^{n-2}x+\binom n{n-4}\cot^{n-4}x-\cdots}$$
$$\implies\cot^n x+n\tan ny\cot^{n-1}x-\binom n2\cot^{n-2}x-\binom n3\tan ny\cot^{n-3}x+\cdots=0$$
So, the roots of $$c^n+n(\tan ny)c^{n-1}-\binom n2c^{n-2}-\binom n3(\tan ny)c^{n-3}+\cdots=0$$ 
are $c_k=\cot\left(y+\dfrac{(2k+1)\pi}{2n}\right), 0\le k\le n-1$
$$\sum _{k=0}^{n-1} \csc ^2\left(\frac{\pi  (2 k+1)}{2 n}+y\right)=\sum _{k=0}^{n-1}\left(1+ \cot^2\left(\frac{\pi  (2 k+1)}{2 n}+y\right)\right)=n+(n\tan ny)^2-2\binom n2=?$$
Similarly, 
$$\csc ^4\left(\frac{\pi  (2 k+1)}{2 n}+y\right)=\left(1+ \cot^2\left(\frac{\pi  (2 k+1)}{2 n}+y\right)\right)^2=?$$
