$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$ It is well-known that one can evaluate the sum 
$$\sum_{k =1}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)=\frac{N^2-1}{3}.$$ 
The answer to this problem can be found here 
click here.
I am now interested in the more difficult problem of evaluating for some $j \in \{1,...,N-1\}$ the sum (we do not sum over $j$)
$$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$$ 
Does this one still allow for an explicit answer?
 A: Start from the well known formula
\begin{equation}
    2^{N-1} \prod _{k=1}^N \left[\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)\right]=\cos N(x-y)-\cos N(x+y),
\end{equation}
take logarithmic derivative
\begin{equation}
\sum _{k=1}^N \frac{\sin(x-y)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)},
\end{equation}
rewrite it as
\begin{equation}
N+\sum _{k=1}^N \frac{\cos(x-y)+\cos \left(x+y+\frac{2\pi k}{N}\right)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{2N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)}\cot(x-y),
\end{equation}
and simplify to get
\begin{equation}
\sum_{k=1}^N\cot\left(x-\frac{\pi k}{N}\right)\cot\left(y-\frac{\pi k}{N}\right)=\frac{2N\cot(x-y)\sin N(x-y)}{\cos N(x-y)-\cos N(x+y)}-N.
\end{equation}
Then take derivatives wrt $x$ and $y$
$$
\sum_{k=1}^N\csc^2\left(x-\frac{\pi k}{N}\right)\csc^2\left(y-\frac{\pi k}{N}\right)=\frac{N}{\sin ^2(x-y)} \left(\frac{N}{\sin ^2Nx}+\frac{N}{\sin ^2Ny}-\frac{2\sin N(x-y)}{\sin Nx\sin Ny}\,\cot (x-y) \right).
$$
A: Yes, it may be simplified. The text below is not probably the shortest way, but it explains how to calculate many similar sums.
We start with $$\sin^2 x=-\frac14(e^{ix}-e^{-ix})^2=-\frac14e^{-2ix}(e^{2ix}-1)^2.$$
So, denoting $\omega_k=e^{2\pi i k/N}$ for $k=0,\ldots,N-1$ we get  $$S:=\sum_{1\leqslant k\leqslant N-1,k\ne j}\csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=
16\sum_{k\ne 0,j}\frac{\omega_k \omega_{k-j}}{(\omega_k-1)^2(\omega_{k-j}-1)^2}=\\
16\omega_j \sum_{k\ne 0,j}\frac{\omega_k^2}{(\omega_k-1)^2(\omega_{k}-\omega_j)^2}.$$
Consider the rational function $f(x):=\frac{x^2}{(x-1)^2(x-a)^2}$ (where $a=\omega_j$). We have
$$f(x)=
\frac{a^2}{(a - 1)^2 (x - a)^2} + \frac{2 a}{(a - 1)^3 (x - 1)} - \frac{2 a}{(a - 1)^3 (x - a)} + \frac1{(a - 1)^2 (x - 1)^2}.$$
Each of four summands may be easily summed up over $x\in \{\omega_0,\ldots,\omega_{N-1}\}\setminus \{1,a\}$. Indeed, we have
$$
\sum_{k} \frac 1{t-\omega_k}=\frac{(t^n-1)'}{(t^n-1)}=\frac{nt^{n-1}}{t^n-1},\,\,\,
\sum_{k} \frac 1{(t-\omega_k)^2}= -\left(\frac{nt^{n-1}}{t^n-1}\right)'=n\frac{t^{n-2}(t^n+n-1)}{(t^n-1)^2}.
$$
If we want to substitute $t=\omega_j$, we get something like
$$
\sum_{k:k\ne j} \frac1{\omega_j-\omega_k}=\left(\frac{nt^{n-1}}{t^n-1}-\frac1{t-\omega_j}\right)_{t=\omega_j}=\left(\frac{(n-1)t^{n}-n\omega_j t^{n-1}+1}{(t^n-1)(t-\omega_j)}\right)_{t=\omega_j}=(n-1)\omega_j^{-1}
$$
by L'Hôpital, analogously for the squares.
