Alternating reciprocal of cosh infinite sum

Today we will prove the following result seen in this post



\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh \pi n} &=\frac{1}{\Gamma^2\left(\frac{3}{4}\right)} \sqrt{\frac{\pi}{2}} 
\end{align*}


Click here to see the proof

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