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

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

$$\begin{gathered} =\frac{-2{\sin }^{2}n\theta +i2\sin n\theta \cos n\theta }{2{\cos }^{2}n\theta +i2\sin n\theta \cos n\theta } \\ =\frac{i^{2}2{\sin }^{2}n\theta +i2\sin n\theta \cos n\theta }{2\cos n\theta \left[\cos n\theta +i\sin n\theta \right]} \\ =\frac{i2\sin n\theta \left[i\sin n\theta +\cos n\theta \right]}{2\cos n\theta \left[\cos n\theta +i\sin n\theta \right]} \\ =i\tan n\theta =R.H.S \end{gathered}$$