An eta Dedekind type product

Prove that

$\prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) = \frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}}$

Solution

Let $\tau = \frac{i}{\sqrt{3}}$ and $q = \exp \left ( 2 \pi i \tau \right )$ . Hence,

$\begin{align*} \prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) &= \prod_{n=1}^{\infty} \left ( 1 + q + q^{2n} \right ) \\ &= \frac{\prod_{n=1}^{\infty} \left ( 1 - q^{3n} \right )}{\prod_{n=1}^{\infty} \left ( 1 - q^n \right )} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta(3\tau)}{\eta(\tau)} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta \left ( - 1/\tau \right )}{\eta(\tau)}\\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \sqrt{-i \tau} \\ &= frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}} \end{align*}$