# Reference request: proof of Ramanujan's Cos/Cosh Identity

The Ramanujan Cos/Cosh Identity, as stated here, is

$$\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\theta}{\cosh n\pi}\right]^{-2}+ \left[1+2\sum_{n=1}^{\infty}\frac{\cosh n\theta}{\cosh n\pi}\right]^{-2}= \frac{2\Gamma^4\left(\frac34\right)}{\pi}.$$

I am looking for a proof, preferably from a reputable source. I hoped I would find something in Ramanujan's Notebooks, but have so far found no mention of it.

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 $$k=k'=1/\sqrt{2}$$.