<s>[INST] $\tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $ 
Prove that $$ \tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $$

There seem to be a lot of similar identities that are provable, for example, by using roots of unity. However, here I cannot get things to work out nicely.
If $u=e^{\frac{2\pi i}{9}} $, then  $$i\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right) =\frac{u-1}{2(u+1)} +2(u^4-u^5)=\frac{-4u^6 +4u^4+u-1}{2(u+1)} $$ and so  $$\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right)^2 = 3 \\ \iff (-4u^6+4u^4+u-1)^2+12(u+1)^2 =0 \\ \iff 16u^8-8u^7+8u^6+8u^5-8u^4+16u^3+13u^2-10u+13 =0$$ Unfortunately, the LHS is not of the form $k(u^8+u^7 +\dots+1)$ making the equality unobvious. How to proceed?
 [/INST] Here's a purely trigonometric solution, in case you're interested:
$$\tan 20°+4\sin 20°=\frac {\sin 20°+2(2\sin 20° \cos 20°)}{\cos 20°}=\frac {(\sin 20°+\sin 40°)+\sin 40°}{\cos 20°}=\frac {2\frac 12 \cos 10°+\cos 50°}{\cos 20°}=\frac {2\cos 30°\cos 20°}{\cos 20°}=2 \frac {\sqrt 3}{2}=\sqrt 3$$
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