Let's start on the RHS.
It is
$$\left[z\prod_{n=1}^\infty\left(1+\frac{iz^2}{2\pi^2 n^2}\right)\right]
\left[z\prod_{n=1}^\infty\left(1-\frac{iz^2}{2\pi^2 n^2}\right)\right].$$
The first bracket here is
$$z\prod_{n=1}^\infty\left(1-\frac{((1-i)z)^2}{4\pi^2 n^2}\right)
=(1+i)\sin\frac{(1-i)z}2$$
and the second is
$$(1-i)\sin\frac{(1+i)z}2.$$
The product is
$$2\sin\frac{(1-i)z}2\sin\frac{(1+i)z}2
=\cos iz-\cos z=\cosh z-\cos z.$$
Start from the infinite product expansion of normalized sinc function:
$${\rm sinc}(z) =\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty \left(1 - \frac{z^2}{n^2}\right)
$$
Substitute $z$ by $2z$, grouping factors in RHS in pairs and then divide it by expansion of ${\rm sinc}(z)$, we obtain an infinite product expansion of $\cos \pi z$:
$$\cos\pi z = \frac{{\rm sinc}(2z)}{{\rm sinc}(z)}
= \prod_{k=1}^\infty\left(1 - \frac{z^2}{(k-\frac12)^2}\right)
$$
Move first factor on RHS to LHS and let $n = k - 1$, this becomes
$$\frac{\cos\pi z}{1 - 4z^2} = \prod_{n=1}^\infty\left(1 - \frac{z^2}{(n+\frac12)^2}\right)\tag{*1}$$
Taking limit at $z = \frac12$ and apply L'Hopital's rule, LHS becomes
$$\lim_{z\to\frac12} \frac{\cos\pi z}{1 - 4z^2} = \lim_{z\to\frac12}
\frac{-\pi \sin\pi z}{-8z} = \frac{\pi}{4}$$
This leads to
$$\frac{\pi}{4} = \prod_{n=1}^\infty\left(1 - \frac{\frac14}{(n+\frac12)^2}\right) = \prod_{n=1}^\infty\frac{n(n+1)}{(n+\frac12)^2}\tag{*2a}$$
In $(*1)$, substitute $z$ by $i\frac{\sqrt{7}}{2}$, we obtain
$$\frac{\cosh(\frac{\pi\sqrt{7}}{2})}{8} = \prod_{n=1}^\infty\left(1 + \frac{\frac74}{(n+\frac12)^2}\right)
= \prod_{n=1}^\infty \frac{n(n+1)+2}{(n+\frac12)^2}\tag{*2b}$$
Divide $(*2a)$ by $(*2b)$, we obtain:
$$\frac{2\pi}{\cosh(\frac{\pi\sqrt{7}}{2})} = \prod_{n=1}^\infty\frac{n(n+1)}{n(n+1)+2} = \prod_{n=1}^\infty \frac{T_n}{T_n + 1}$$
Best Answer
As already noticed in comments, the expression can be obtained from the infinite products for $\Gamma$ (either Euler's one, or Weierstrass's one): $$\Gamma(1+z)=\prod_{n=1}^\infty\frac{(1+1/n)^z}{1+z/n}=e^{-\gamma z}\prod_{n=1}^\infty\frac{e^{z/n}}{1+z/n},$$ and the "algebraic" $1-x/n^3=(1-x^{1/3}/n)(1-e^{2\pi i/3}x^{1/3}/n)(1-e^{-2\pi i/3}x^{1/3}/n)$, giving $$\prod_{n=1}^\infty\left(1-\frac{x}{n^3}\right)=\Big(\Gamma(1-x^{1/3})\Gamma(1-e^{2\pi i/3}x^{1/3})\Gamma(1-e^{-2\pi i/3}x^{1/3})\Big)^{-1}.$$ This easily applies to more general "rational infinite products", as outlined here.