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

I think the best solution here is using right triangle . . . I have one too, but not pretty.
 [/INST] Multiplying by $\cos(\pi/9)$, this is equivalent to
$$\sin\frac\pi 9+2\sin\frac{2\pi}9=\sqrt3\cos\frac\pi 9.\tag1$$
But
$$\sqrt3\cos\frac\pi 9-\sin\frac\pi9=
2\left(\sin\frac\pi3\cos\frac\pi 9-\cos\frac\pi3\sin\frac\pi9\right)
=2\sin\left(\frac\pi3-\frac\pi9\right)=2\sin\frac{2\pi}9$$
which proves $(1)$.
</s>