$ \sum_{n=1}^\infty \csc^2(\omega\pi n)= \frac{A}{\pi} +B $

In my mind, this is most naturally linked to Eisenstein series.

For $\tau\in\mathbb{C}$ with $\Im\tau>0$, and an integer $k>1$, these are defined by $$G_{2k}(\tau)=\sum_{m,n\in\mathbb{Z}}'(m+n\tau)^{-2k}:=\sum_{(m,n)\in\mathbb{Z}^2\setminus\{(0,0)\}}(m+n\tau)^{-2k}$$ (i.e., $\sum'$ means $\sum$ over $(m,n)$ excluding $m=n=0$), and satisfy the well-known $G_{2k}(\tau+1)=G_{2k}(\tau)$ and $G_{2k}(-1/\tau)=\tau^{2k}G_{2k}(\tau)$ (the latter equality corresponds to the validity of $\sum_m\sum_n=\sum_n\sum_m$).

For $k=1$, the series doesn't converge in the usual sense; however, it's possible to define $$G_2(\tau)=\sum_{n\in\mathbb{Z}}\sum_{m\in\mathbb{Z}}'(m+n\tau)^{-2}$$ where, in the $\sum'$, the term with $m=0$ is excluded if $n=0$. We still have $G_2(\tau+1)=G_2(\tau)$ but $$G_2(-1/\tau)=\tau^2 G_2(\tau)-2i\pi\tau$$ (now the order of the summations is important).

To prove the last formula, one usually "approximates" $G_2(\tau)$ with $G(\tau)=\sum_n\sum_m'\frac{1}{(m+n\tau)(m+1+n\tau)}$ (here, the meaning of $\sum'$ is clear but obviously different!), so that $G_2(\tau)-G(\tau)$ is absolutely convergent (hence the interchange of summations is valid), and both $G(\tau)$ and its version with the summations interchanged are easy to evaluate in closed form (because of "telescoping").

The link to the present question is given by $\pi^2\csc^2\pi\tau=\sum_{m\in\mathbb{Z}}(m+\tau)^{-2}$ (see (1), (2), etc.): $$\sum_{n=1}^\infty\csc^2n\pi\omega=\frac{1}{\pi^2}\sum_{n=1}^\infty\sum_{m\in\mathbb{Z}}(m+n\omega)^{-2}=\frac{1}{2\pi^2}\left(G_2(\omega)-\sum_{m\neq 0}\frac{1}{m^2}\right)=\frac{\color{blue}{G_2(\omega)}}{2\pi^2}-\frac16;\\G_2(\omega)=G_2(1+\omega)=G_2(-1/\omega)=\omega^2 G_2(\omega)-2i\pi\omega\implies \color{blue}{G_2(\omega)=2\pi/\sqrt{3}}.$$