If you are able to show that
$$ \sum_{n\geq 0}\frac{(-1)^n}{a^2+n^2} = \frac{1}{2a^2}+\frac{\pi}{2a\sinh(\pi a)} \tag{1}$$
$$ \sum_{n\geq 0}\frac{(-1)^n}{a^2+(2n)^2} = \frac{1}{2a^2}+\frac{\pi}{4a\sinh(\pi a/2)} \tag{2}$$
for instance through the Fourier sine series of $\sinh$ over $(-\pi,\pi)$, then by considering the difference between $(1)$ and $(2)$:
$$ \sum_{n\geq 0}\frac{\color{red}{1}}{a^2+(2n+1)^2} = \frac{\pi\tanh(\pi a/2)}{4a} \tag{3}$$
but if the last $\color{red}{1}$ is replaced by $(-1)^n$ the outcome is not as elementary. We have
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(s)-\psi(1-s)=-\pi\cot(\pi s)\tag{4} $$
where $\psi(s)=\frac{d}{ds}\log\Gamma(s)$, hence
$$ \sum_{n\geq 0}\frac{\color{red}{(-1)^n}}{a^2+(2n+1)^2}=\frac{1}{4a}\,\text{Im}\left[\psi\left(\tfrac{1+ia}{4}\right)-\psi\left(\tfrac{3+ia}{4}\right)\right]\tag{5}$$
which essentially has a nice closed form (Catalan's constant $G$) only for $a\to 0^+$.