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I expand my comment into an answer.

The key here is the Fourier series for the elliptic function $\operatorname {dn} (u, k) $ given as $$\operatorname {dn} (u, k) =\frac{\pi} {2K}\left(1+4\sum_{n=1}^{\infty} \frac{q^n} {1+q^{2n}}\cos\left(\frac{n\pi u} {K} \right) \right) $$ where $K$ is the complete elliptic integral of first kind corresponding to modulus $k$ and $q=\exp(-\pi K'/K) $ is the nome corresponding to modulus $k$.

Let $\theta=\pi u/K$ then we have $$\operatorname {dn} \left(\frac{K\theta} {\pi}, k\right) =\frac{\pi} {2K}\left(1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh (n\pi K'/K)} \right) $$ Putting $K'=K$ so that $k=1/\sqrt{2}$ and $K=\dfrac{\Gamma^2(1/4)}{4\sqrt{\pi}}$ we get $$1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh n\pi} =\frac{2K}{\pi}\operatorname {dn} \left(\frac{K\theta} {\pi}, \frac{1}{\sqrt{2}}\right)$$ Let the above expression be denote by $A$ and the expression obtained from it by replacing $\theta$ with $i\theta$ be denoted by $B$. Then we have to show that $$\frac{1}{A^2}+\frac{1}{B^2}=\frac{2\Gamma^{4}(3/4)}{\pi}$$ Note that we have $$\Gamma(1/4)\Gamma (3/4)=\sqrt{2}\pi$$ and therefore $$\frac{2K} {\pi} =\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}$$ and we have then $$A^{-2}+B^{-2}=\frac{\Gamma ^{4}(3/4)}{\pi}\left(\operatorname {dn} ^{-2}\left(\frac{K\theta}{\pi}\right)+\operatorname {dn} ^{-2}\left(\frac{iK\theta}{\pi}\right)\right)$$ The expression in parentheses is easily seen to be $2$ if we note that $$\operatorname {dn} (iu, k) =\frac{\operatorname {dn} (u, k')} {\operatorname {cn} (u, k')} $$ and here $k=k'=1/\sqrt{2}$.