<s>[INST] Evaluating $\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$ 
How to evaluate this?
$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}$$

My work
I couldn't figure out a method to solve this product.
I thought that this identity could help.
$$\frac{e^{i\theta}-1}{e^{i\theta}+1}=i\tan \frac{\theta}{2}$$
By supposing $z=e^{\frac{2k\pi}{2m+1}i}$, then,
$$i\tan \frac{k\pi}{2m+1}=\frac{z-1}{z+1}$$
So, the product
$$\displaystyle\prod_{k=1}^m \tan\frac{k\pi}{2m+1}=\displaystyle\prod_{k=1}^m \frac{z-1}{i(z+1)}=\displaystyle\prod_{k=1}^m \frac{e^{\frac{2k\pi}{2m+1}i}-1}{i(e^{\frac{2k\pi}{2m+1}i}+1)}$$
which is getting more complicated.
Answer is $\sqrt{2m+1}$. Any help is appreciated.
 [/INST] Let
$$f(x) = x^{n-1} + x^{n-2} + x^{n-3} \>\cdots \> +\>x +1
= \prod_{k=1}^{n-1}(x - e^{i\frac{2\pi k}n})
$$
and set $n=2m+1$ to evaluate
$$\frac{f(1)}{f(-1)}= 2m+1 
= \prod_{k=1}^{2m}\frac{1 - e^{i\frac{2\pi k}{2m+1}}}{1 + e^{i\frac{2\pi k}{2m+1}}}
=  i^{2m} \prod_{k=1}^{2m}\tan \frac{k\pi}{2m+1}
= \prod_{k=1}^{m}\tan^2\frac{k\pi}{2m+1}
$$
Thus
$$\prod_{k=1}^m \tan \frac{k\pi}{2m+1}=\sqrt{2m+1}$$
</s>