We're given
[tex]z = \mathrm{cis}(\theta) = \cos(\theta) + i \sin(\theta)[/tex]
By de Moivre's theorem,
[tex]z^n = \cos(n\theta) + i \sin(n\theta)[/tex]
so that
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{\cos(2\theta) + i \sin(2\theta) - 1}{\cos(2\theta) + i \sin(2\theta) + 1}[/tex]
Multiply uniformly by the conjugate of the denominator.
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{\cos(2\theta) + i \sin(2\theta) - 1}{\cos(2\theta) + i \sin(2\theta) + 1}\cdot\dfrac{\cos(2\theta) - i \sin(2\theta) + 1}{\cos(2\theta) - i \sin(2\theta) + 1}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{\cos^2(2\theta) - 1 + 2i\sin(2\theta) + \sin^2(2\theta)}{(\cos(2\theta)+1)^2 + \sin^2(2\theta)}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{2i\sin(2\theta)}{\cos^2(2\theta) + 2\cos(2\theta) + 1+ \sin^2(2\theta)}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{2i\sin(2\theta)}{2\cos(2\theta) + 2}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{i\sin(2\theta)}{\cos(2\theta) + 1}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{2i\sin(\theta)\cos(\theta)}{2\cos^2(\theta) - 1 + 1}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = \dfrac{i\sin(\theta)}{\cos(\theta)}[/tex]
[tex]\dfrac{z^2 - 1}{z^2 + 1} = i\tan(\theta)[/tex]
QED
