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.

$$(\tan x+ 4\sin x)^{2}- 3= - \frac{1}{4}(2\cos x+ 1)(2\cos 3x- 1)\csc^{2}\left ( \frac{\pi}{4}- \frac{x}{2} \right )\csc^{2}\left ( \frac{x}{2}+ \frac{\pi}{4} \right )$$
Therefore
$$\tan\frac{\pi}{9}+ 4\sin\frac{\pi}{9}= \sqrt{3}$$
