Product of Sines and Sums of Squares of Tangents

Let us define $\theta=\frac{\pi}{2n+1}$ and $~\xi_k=\tan(k\theta)$ for $-n\le k\le n$. Clearly $$\xi_k=\frac{1}{i}\frac{e^{2ik\theta}-1}{e^{2ik\theta}+1}\iff\frac{1+ i\xi_k}{1-i\xi_k}=e^{2ik\theta}$$ and since $e^{i2(2n+1)\theta}=1$ we conclude that $\{\xi_k:-n\le k\le n\}$ are zeros of the polynomial $P(X)$ defined by $P(X)=(X+i)^{2n+1}+(X-i)^{2n+1}$, but $\deg P=2n+1$ and the coefficient of $X^{2n+1}$ in $P$ is $2$. So $$P(X)=2\prod_{k=-n}^n(X-\xi_k)=2X \prod_{k=1}^n(X^2-\xi_k^2)$$ because $\xi_{-k}=-\xi_k$. On the other hand $$\eqalign{P(X)&=\sum_{\ell=0}^{2n+1}\binom{2n+1}{\ell}X^\ell i^{2n+1-\ell}(1-(-1)^{\ell})\cr &=2X\sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}X^{2\ell} (-1)^{n-\ell}\cr }$$ It follows that $$\prod_{k=1}^n(X^2-\xi_k^2)= \sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}X^{2\ell} (-1)^{n-\ell}$$ Or $$\prod_{k=1}^n(t-\xi_k^2)= \sum_{\ell=0}^{n}\binom{2n+1}{2\ell+1}t^{\ell} (-1)^{n-\ell}\tag{1}$$ Now comparing the coefficients of $t^{n-1}$ on both sides we get $$\sum_{k=1}^n\xi_k^2=\binom{2n+1}{2n-1}=\binom{2n+1}{2}=n(2n+1)$$ or $$\sum_{k=1}^n\tan^2\frac{\pi k}{2n+1}=n(2n+1)\tag{2}$$ Again, comparing the constant terms in $(1)$ we get $$\prod_{k=1}^n\xi_k^2=\binom{2n+1}{1}=2n+1$$ and since $\prod_{k=1}^n\xi_k>0$ we get $$\prod_{k=1}^n\tan\frac{\pi k}{2n+1}=\sqrt{2n+1}\tag{3}$$ This implies also that $$\prod_{k=1}^n\sin\frac{\pi k}{2n+1}= \prod_{k=1}^n\tan\frac{\pi k}{2n+1}\prod_{k=1}^n\cos\frac{\pi k}{2n+1}= \frac{\sqrt{2n+1}}{2^n}\tag{4}$$ and we are done.$\qquad\square$