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.
Best Answer
Your approach will work with double angle formulae, but this is quicker: since $z=\exp i\theta$, $\frac{z-1/z}{z+1/z}=\frac{2i\sin\theta}{2\cos\theta}$.