The identity $$ \sum_{n=1}^{\infty}\chi(n)\arctan e^{-\alpha n}+\sum_{n=1}^{\infty}\chi(n)\arctan e^{-\beta n}=\frac{\pi}{8}, \qquad \alpha\beta=\frac{\pi^2}{4},\tag{1} $$ where $\chi(n)=\sin\frac{\pi n}{2}$ is Dirichlet character modulo $4$, known from the theory of elliptic functions (Ramanujan's Notebooks, part II, ch.14, entry 15) has a finite analog \begin{align} \sum_{|j|\le n}&(-1)^{n+j}\arctan {\biggl(\!\sqrt{1+\alpha^2\cos^2\!\tfrac{\pi j}{2n+1}}-\alpha\cos\tfrac{\pi j}{2n+1}\!\biggr)^{2 m+1}}\\&+\sum_{|k|\le m}(-1)^{m+k}\arctan\!{\biggl(\!\sqrt{1+\beta^2\cos ^2\!\tfrac{\pi k}{2m+1}}-\beta\cos\tfrac{\pi k}{2m+1}\!\biggr)^{2 n+1}}=\frac{\pi}{4},\tag{1a} \end{align} where $n,m\in\mathbb{N}_0$ and $\alpha\beta=1$, $\alpha>0$. Namely, $(1)$ follows from $(1a)$ when $n=m\to\infty$, after suitable redefinition of $\alpha$ and $\beta$. (for details see https://arxiv.org/abs/2003.05306 )
The following identity was found in an unpublished manuscript by B. Cais, On the transformation of infinite series, (1999): \begin{align} \sum_{n=1}^{\infty} \left(\frac{n}{3}\right)&\arctan\frac{\sqrt{3}}{1+2 e^{\alpha n}}\\&+\sum _{n=1}^{\infty} \left(\frac{n}{3}\right) \arctan\frac{\sqrt{3}}{1+2 e^{\beta n}}=\frac{\pi}{18}, \qquad \alpha\beta=\frac{4\pi^2}{9},\tag{2} \end{align} where $\left(\frac{j}{3}\right)$ is Legendre symbol.
In the arxiv preprint cited above, it was found that \begin{align} \sum_{n=0}^{\infty} \left(\frac{n-1}{3}\right)&\arctan\frac{\sqrt{3}}{1-2 e^{\alpha (2n+1)}}\\&+\sum _{n=0}^{\infty} \left(\frac{n-1}{3}\right) \arctan\frac{\sqrt{3}}{1-2 e^{\beta (2n+1)}}=\frac{2\pi}{9}, \qquad \alpha\beta=\frac{\pi^2}{9}.\tag{3} \end{align}
Q: Do finite analogs of $(2)$ and $(3)$ exist and how to find them?
A finite analog of a certain linear combination of $(2)$ and $(3)$ have been found in the arxiv preprint cited above. But no finite analog for $(2)$ or $(3)$ separately is known. It is not known how to extend the method used in the preprint to answer this question. Any ideas and comments are welcome.