Is $\prod_{n=0}^\infty \left(1-\frac{1}{\cosh ^2((n+1/2)\pi)}\right)=\frac{1}{\sqrt[4]{2}}$ true? The infinite product
$$\prod_{n=0}^\infty \left(1-\frac{1}{\cosh  ^2((n+1/2)\pi)}\right)$$
agrees with $\frac{1}{\sqrt[4]{2}}$ to at least 100 decimal places.
The "identity" is reminiscent of
$$\sqrt[4]{1-\lambda (i)}=\frac{1}{\sqrt[4]{2}}$$
where $\lambda$ is the modular lambda function. I tried to use
$$\theta_3(z|\tau)=\theta_3(0,\tau)\prod_{n=0}^\infty \frac{\cos ((n+1/2)\pi\tau+z)\cos ((n+1/2)\pi\tau -z)}{\cos^2 ((n+1/2)\pi \tau)}$$
where $z,\tau\in\mathbb{C}$ and $\operatorname{Im}\tau\gt 0$.
$\theta_3(\pi/2|i)$ leads to
$$\prod_{n=0}^\infty \tanh^2 \left(\left(n+\frac{1}{2}\right)\pi\right)$$
and $\theta_3(i|i)$ leads to
$$\prod_{n=0}^\infty \cosh \left(1-\left(n+\frac{1}{2}\right)\pi\right)\cosh\left(1+\left(n+\frac{1}{2}\right)\pi\right)\operatorname{sech}^2\left(\left(n+\frac{1}{2}\right)\pi\right)$$
but these products don't seem to give the answer.
 A: If $q=e^{-\pi} $ then we can see that your product equals $$\prod_{n\geq 1}\frac{(1-q^{2n-1})^2}{(1+q^{2n-1})^2}$$ which equals $(g_1/G_1)^2$ where $g, G$ represent Ramanujan class invariants. This is indeed $2^{-1/4} $ as $G_1=1,g_1=2^{-1/8}$.

A little amount of explanation about Ramanujan class invariants is necessary here.
Let $k\in(0,1)$ be the elliptic modulus and define complete elliptic integral of first kind $$K(k) =\int_{0}^{\pi/2}\frac{dx} {\sqrt{1-k^2\sin^2x}}\tag{1}$$ Further we define complementary modulus $k'=\sqrt{1-k^2}$ and the expressions $K(k), K(k') $ are usually denoted by $K, K'$ if the value of $k$ is known from context.
A lot of magic is hidden in the elliptic modulus $k$ and the value of $k$ can be obtained if $K, K'$ are given using nome $q=\exp (-\pi K'/K) $.
We have $$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}, k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}\tag{2}$$ where
\begin{align} 
\vartheta_2(q)&=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2}\notag\\
&=2q^{1/4}\prod_{n=1}^{\infty}(1-q^{2n})(1+q^{2n})^2 \tag{3a}\\
\vartheta_3(q)&=\sum_{n\in\mathbb {Z}} q^{n^2}\notag\\
&=\prod_{n=1}^{\infty} (1-q^{2n})(1+q^{2n-1})^2\tag{3b}\\
\vartheta_4(q) &= \vartheta_3 (-q) \tag{3c}
\end{align}
are theta functions of Jacobi with one parameter being $0$. The equality of series and product expressions above is due to Jacobi Triple Product identity.
Ramanujan defined his class invariants $g_N, G_N $ using functions $g, G$ as \begin{align}
g(q) &=2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty} (1-q^{2n-1})\tag{4a}\\
G(q)&=2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})\tag{4b}\\
g_N &=g(\exp(-\pi\sqrt{N}))\tag{4c}\\
G_N &=G(\exp (-\pi\sqrt{N})) \tag{4d}
\end{align}
where $N$ is a positive rational number.
It can be proved using the product expressions for theta functions that $$g(q) =(2k/k'^2)^{-1/12},G(q)=(2kk')^{-1/12}\tag{5}$$ It can also be proved with some effort (say using the theory of modular equations)  that if $N$ is a positive rational number and $q=\exp(-\pi\sqrt{N}) $ then the values of $k, k'$ are algebraic and hence $G_N, g_N $ are also algebraic numbers.
If $N=1$ we have $$q=e^{-\pi} =\exp (-\pi K'/K) $$ so that $K'=K$ and $k'=k=1/\sqrt{2} $ and from $(5)$ we get $g_1=2^{-1/8},G_1=1$. Using these values the product in question evaluates to $2^{-1/4}$.
