Conjecture on limit of a partial fraction expansion involving tan and cot

Question

Let $q$ be an integer such that $q \equiv 1\mod{4}$ or $q \equiv 2\mod{4}$; i.e., q={... -7,-6,-3,-2,1,2,5,6,9...} I seek a proof of the following conjecture: $$(C)\quad \lim_{n \to \infty} \sum_{k=1}^n \frac{\tan^2\big(\dfrac{k \pi}{4n+1}\big)}{x+\tan^2\big(\dfrac{k \pi}{4n+1}\big) } -\frac{\cot^2\big(\dfrac{(k+q\,n) \pi}{4n+1}\big)}{x+\cot^2\big(\dfrac{(k+q\,n) \pi}{4n+1}\big) } = (-1)^q\,\frac{(2 \lfloor (q-2)/4 \rfloor+1)}{2(x+1)}$$ The conjectured limit came about from a long series of generalizations from the identity, $$\frac{8}{\pi}\int_0^{\pi/2} u\,\tan^{-1}(\tan^{4n+1} u)du - \frac{\pi^2}{3}=$$ $$ \sum_{k=1}^n \text{Li}_2 \Big(-\cot^2 \big(\dfrac{(k+n) \pi}{4n+1}\big)\Big) - \text{Li}_2 \Big(-\tan^2 \big(\dfrac{k \pi}{4n+1}\big)\Big).$$ To see the relationship with $(C),$ if within the integral the limit $n \to \infty$ is taken, the 'trig part' degenerates to a step function in which the integral can easily be performed. The dilogarithms can be written as an integral, and the integrand must obey (C), but this only holds for $q=1.$ The intermediate conjecture was, in fact, $$(C')\quad \lim_{n \to \infty} \sum_{k=1}^n \text{Li}_s \Big(-\cot^2 \big(\dfrac{(k+q\,n) \pi}{4n+1}\big)\Big) - \text{Li}_s \Big(-\tan^2 \big(\dfrac{k \pi}{4n+1}\big)\Big)$$ $$= (-1)^{q+1}\, \frac{\text{Li}_s(-1)}{2} (2 \lfloor (q-2)/4 \rfloor+1), \quad \frac{\Gamma(s)\,\text{Li}_s(r)}{r}=\int_0^\infty\frac{t^{s-1}}{e^t-r}dt$$ Express the polylogs as the given integral and then examine the integrands to see how $(C')$ implies $(C).$ I don't expect the polylog to have some undiscovered property so it is natural to suspect that $(C)$ is the more fundamental conjecture. I have numerically checked $(C')$ for n=5000 and for the set of $q$ given at the top of this question (without ellipses) and for $s=2,3,5/2,$ and $8/3.$ I have checked $(C)$ for many $x,q,$ and $n$ as well.