How to derive Weierstrass factorization for $\cos(z)$ from $\tan(z)=\sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2-4z^2}$

Question

Given that: $\tan(z)=\sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2-4z^2}$

Integrate on both sides:

$-\ln(\cos(z))=-\sum_{n=0}^\infty \left( \ln\left[ (2n+1)^2-4z^2 \right]+\ln c_n\right)$

Simplify:

$\cos(z)=\prod_{n=0}^\infty c_n \left[ (2n+1)^2-4z^2 \right]$

Determine $c_n$:

let $z=0$

$1=\prod_{n=0}^\infty c_n(2n+1)^2$

One choice (also, this is the correct choice) is:

$c_n=\frac{1}{(2n+1)^2}$

But how to show this choice is unique?