> 
> Prove that, for a natural number $n$,
>  $$\prod\_{k=1}^n \tan\left(\frac{k\pi}{2n+1}\right) = 2^n \prod\_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$
> 
> 
> 

This follows from a continued fraction identity for which, I think, there is a lengthy proof. But, I thought, that there may be a direct geometric or another proof. Constructing a polynomial with the sines and tangents roots may be helpful.