<s>[INST] If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$ 
If $z=\cos\theta + i\sin \theta$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$

Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
$$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$
$$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$
$$\frac{\cos2\theta + i\sin 2\theta  -1}{\cos2\theta + i\sin 2\theta +1}$$
I understand how I can do it with using $z=e^{i \theta}$, however I want to solve it using double angle identities.  
 [/INST] Now where you've left off,
use $\cos2y=2\cos^2y-1=1-2\sin^2y\ \ \ \  (1)$
to find $$\frac{\cos2\theta + i\sin 2\theta  -1}{\cos2\theta + i\sin 2\theta +1}=\dfrac{-2\sin^2\theta+ i\sin 2\theta}{2\cos^2\theta+ i\sin 2\theta}=\dfrac{2i\sin\theta(\cos\theta+i\sin\theta)}{2\cos\theta(\cos\theta+i\sin\theta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta $$
 use $(1)$ to establish $$\dfrac{z^n-1}{z^n+1}=2i\tan\dfrac{n\theta}2$$
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