show that:$$\tan^4{\dfrac{\pi}{11}}+\tan^4{\dfrac{2\pi}{11}}+\tan^4{\dfrac{3\pi}{11}}+\tan^4{\dfrac{4\pi}{11}}+\tan^4{\dfrac{5\pi}{11}}=2365$$
my try:
I think first we can find this value：$$x=\tan{\dfrac{\pi}{11}}+\tan{\dfrac{2\pi}{11}}+\tan{\dfrac{3\pi}{11}}+\tan{\dfrac{4\pi}{11}}+\tan{\dfrac{5\pi}{11}}$$ But I can't.Thank you for you help.
Using this for odd $n=2m+1$ and setting $\tan(2m+1)x=0\implies x=\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
So, the roots of $$\tan^{2m+1}x-\binom{2m+1}2\tan^{2m-1}x+\binom{2m+1}4\tan^{2m-3}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $0\le r\le 2m$
Discarding $\tan 0=0,$ the roots of $$\tan^{2m}x-\binom{2m+1}2\tan^{2m-2}x+\binom{2m+1}4\tan^{2m-4}x-\cdots=0$$ are $\tan\frac{r\pi}{2m+1}$ where $1\le r\le 2m$
Now observe that $\displaystyle \tan\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan\left(\pi-\frac{u\pi}{2m+1}\right)=-\tan\left(\frac{u\pi}{2m+1}\right)$
$\displaystyle\implies\tan^2\left(\frac{(2m+1-u)\pi}{2m+1}\right)=\tan^2\left(\frac{u\pi}{2m+1}\right)$
So, $\displaystyle\tan^2\frac{r\pi}{2m+1}$ where $1\le r\le m$ or $m+1\le r\le2m$ (more generally we can replace $u,1\le u\le m$ with $2m+1-u$) are the $m$ roots of $$t^mx-\binom{2m+1}2t^{m-1}x+\binom{2m+1}4t^{m-2}x-\cdots=0$$
$\displaystyle\sum_{1\le r\le m}\tan^4\frac{r\pi}{2m+1}= \left(\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}\right)^2-2\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}$ where $1\le i<j\le m$
Now using Vieta's formula,
$\displaystyle\sum_{1\le r\le m}\tan^2\frac{r\pi}{2m+1}=\binom{2m+1}2$
and $\displaystyle\sum_{1\le r\le m}\tan^2\frac{r_i\pi}{2m+1}\tan^2\frac{r_j\pi}{2m+1}=\binom{2m+1}4$ where $1\le i<j\le m$
In this problem $m=5$
