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.  

If you divide the numerator and denominator of
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
by $\cos^2\theta$ and then use the identity $\sec^2\theta=1+\tan^2$ on the result you obtain
$$ \frac{(1+i\,\tan\theta)^2-1-\tan^2\theta}{(1+i\,\tan\theta)^2+1+\tan^2\theta} $$
which simplifies to
$$ \frac{i\,\tan\theta-\tan^2\theta}{1+i\,\tan\theta} $$
Then factor the numerator to get
$$ \frac{i\,\tan\theta(1+i\,\tan\theta)}{1+i\,\tan\theta}=i\,\tan\theta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i\,\theta}=\cos\theta+i\,\sin\theta$
