<s>[INST] $\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$ 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?
 [/INST] 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 $y$
$$
\sum_{k=1}^N\csc^2\left(x-\frac{\pi k}{N}\right)\csc^2\left(y-\frac{\pi k}{N}\right)=\frac{N}{\sin ^2(x-y)} \left(\frac{N}{\sin ^2Nx}+\frac{N}{\sin ^2Ny}-\frac{2\sin N(x-y)}{\sin Nx\sin Ny}\,\cot (x-y) \right).
$$
</s>