What I'd like to find is an identity for $$\sum_{k=0}^{n-1}\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)$$
here it can be shown that where $x=y$,
$$n^2 \csc^2(nx) = \sum_{k=0}^{n-1}\csc^2\left(x+ k \frac{\pi}{n}\right)$$
I also looked at possible residue methods involving csc
I thought I could use
$\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 +\sum^{n-1}_{j=0}\sum^{n-1}_{i\neq j} Z_jZ_i$
but I have had no luck so far.
Best Answer
$$\bbox[5pt,border:2pt solid black]{\sum_{k=0}^{n-1}\csc\left(x+\frac{k\pi}n\right)\csc\left(y+\frac{k\pi}n\right)=\frac{n\sin n(x-y)}{\sin(x-y)\sin nx\sin ny}}$$
Steps to get it: $\cos nx$ is a polynomial in $\cos x$, then $\cos nx-\cos ny=F(\cos x)$ may be considered as a polynomial in $\cos x$ (with coefficients depending on $y$). Assume (temporarily) that $y$ is not a rational multiple of $\pi$; then the roots of $F(z)$ are given by $z_k=\cos(y+2k\pi/n)$ for $0\leqslant k\leqslant n-1$. Hence $F(z)=A\prod_{k=0}^{n-1}(z-z_k)$, with $A$ depending on $y$ only (in fact $A=2^{n-1}$ but we don't need it here; also, now the assumption on $y$ can be dropped by continuity). If we put $z=\cos x$ back, and take the logarithmic derivative w.r.t. $x$, we get $$\frac{n\sin nx}{\cos nx-\cos ny}=\sum_{k=0}^{n-1}\frac{\sin x}{\cos x-\cos(y+2k\pi/n)}.$$
To get the formula at the beginning, put $(x,y)\gets(x-y,x+y)$.