# Proving $\tan\frac{4\pi}{11} + 4\sin\frac{\pi}{11} = \sqrt{11}$

In a similar vein as $$\tan\frac{3\pi}{11} + 4\sin\frac{2\pi}{11} = \sqrt{11}$$ discussed in this question is this identity:

$$\tan\frac{4\pi}{11} + 4\sin\frac{\pi}{11} = \sqrt{11}$$

Trying to adopt a method on the same line however lamentably fails. I wonder if the arguement of $$11$$th roots of unity can still be effectively employed in this case. Is there a way to adapt it or there could be possibly an easier way out to prove the result?

Let $$a=\frac\pi{11}$$ to evaluate
\begin{align} & 4 (\sin 2a -\sin a)-(\tan 4a - \tan3a )\\ = & 4\sin a (2\cos a -1)-\frac{\sin a}{\cos 3a\cos4a} =\frac{4\sin a}{\cos 7a+\cos a}\cdot A\tag1 \end{align}
where \begin{align} A = & 2\cos a( \cos7a + \cos a) - (\cos7a + \cos a)-\frac12 \\ = & \cos10a+ \cos8a+ \cos 6a +\cos 4a +\cos2a +\frac12\\ = & \frac12 \sum_{k=0}^{10} e^{i 2ka}=0 \\ \end{align}
Substitute $$A=0$$ into (1) to obtain
$$\tan\frac{4π}{11} + 4\sin\frac{π}{11} = \tan\frac{3π}{11} + 4\sin\frac{2π}{11}= \sqrt{11}$$