If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\cos ^{2} \theta}\right| \\ =\left|\sqrt{1-\left(\frac{z^{2}+1}{2z}\right)^{2}}\right| \\ =\left|\sqrt{\frac{4 z^{2}-z^{4}-2 z^{2}-1}{4 z^{2}}} \right|\\ =\left|\sqrt{\frac{-\left(z^{4}-2 z^{2}+1\right)}{4 z^{2}}}\right|=\left|\sqrt{\frac{\left(z^{2}-1\right)^{2}}{4 z^{2}}}\right| \\ =|\frac{z^{2}-1}{2 z}|\end{array}
$$
So $|\tan \theta|=\left|\frac{z^{2}-1}{z^{2}+1}\right|$ ( proven when $n = 1$)
Is there any way to prove directly by taking $n$?

$$z+\frac 1z=2\cos \theta\Rightarrow z^2-2\cos\theta z+1=0$$
Solve this quadratic to obtain 
$$z=cos\theta\pm i\sin\theta$$
Or,
$$z=e^{i\theta},e^{-i\theta}$$
Now,
$$\frac{z^{2n}-1}{z^{2n}+1}=\frac{z^n-1/z^n}{z^n+1/z^n}$$
In either of the above cases(for $z$), we have,
$$z^n+1/z^n=e^{in\theta}+e^{-in\theta}=2\cos n\theta$$
But depending on the value of $z$ we have
$$z^n-1/z^n=\pm \left(e^{in\theta}-e^{-in\theta}\right)=\pm 2\sin n\theta$$
We now have
$$\left |\frac{z^{2n}-1}{z^{2n}+1}\right |=\left|\frac{\pm 2\sin n\theta}{2\cos n\theta}\right|=|\tan n\theta|$$
Hence proved.
