<s>[INST] 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?
 [/INST] 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
$$\tan\frac{4π}{11} + 4\sin\frac{π}{11} = \tan\frac{3π}{11} + 4\sin\frac{2π}{11}= \sqrt{11}$$
where How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
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