If $\text{'}n\text{'}$ is an integer and $z=cis\theta , \left(\theta \ne \left(2n+1\right)\frac{\pi }{2}\right)$ then show that $\frac{z^{2n}-1}{z^{2n}+1}=i \tan n\theta$

Given that $Z=e^{i\theta }$

$$\begin{gathered} Z^{2n}={\left(e^{i\theta }\right)}^{2n}=e^{2n\theta i}=\cos n\theta +i\sin n\theta \\ {Z}^{2n}-1=\left[\cos 2n\theta +i\sin n\theta \right]-1 \\ =-2{\sin }^{2}n\theta +2i\sin n\theta .\cos n\theta \\ =2i\sin n\theta \left(\cos n\theta +i\sin n\theta \right) ...\left(1\right) \\ Z^{2n}+1=2\cos n\theta \left[\cos n\theta +i\sin n\theta \right] ...\left(2\right) \\ L.H.S=\frac{Z^{2n}-1}{ Z^{2n}+1}=\frac{2i \sin n\theta }{2\cos n\theta }=i \tan n\theta =R.H.S \end{gathered}$$