Show that
$ \displaystyle \prod_{n=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} = \displaystyle \prod_{n=2}^{\infty}\frac{(k-1)(k+1)}{k^{2}+1} =\frac{\pi}{\sinh \pi}$
We know the product representation
$$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$
it is rather easy.
Setting $z = i$, we obtain
$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$
On the other hand,
$$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$
so
$$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$
Now divide.
$ \displaystyle \prod_{n=2}^{\infty}\frac{k^{2}-1}{k^{2}+1} = \displaystyle \prod_{n=2}^{\infty}\frac{(k-1)(k+1)}{k^{2}+1} =\frac{\pi}{\sinh \pi}$
Solution:
We know the product representation
$$\sin (\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \frac{z^2}{k^2}\right),\tag{1}$$
it is rather easy.
Setting $z = i$, we obtain
$$\frac{\sin (\pi i)}{\pi i} = \frac{\sinh \pi}{\pi} = \prod_{k=1}^\infty \left(1 - \frac{i^2}{k^2}\right) = \prod_{k=1}^\infty \left(1 + \frac{1}{k^2}\right) = 2 \prod_{k=2}^\infty \left(\frac{k^2+1}{k^2}\right).$$
On the other hand,
$$\prod_{k=2}^n \left(\frac{k^2-1}{k^2}\right) = \frac12\cdot \frac32\cdot \frac23\cdot \frac43 \dotsb \frac{n-1}{n}\cdot \frac{n+1}{n} = \frac12\cdot\frac{n+1}{n},$$
so
$$\prod_{k=2}^\infty \left(\frac{k^2-1}{k^2}\right) = \frac12.$$
Now divide.
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