<s>[INST] Prove that $4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$ . 
Prove that $$4\sin\frac{2\pi}{7}- \tan\frac{\pi}{7}= \sqrt{7}$$

I think midway calculations are not easy formulas, I can't find what kind of formula transformation to solve! I think the best solution here is using right triangle . . . I have one solution too, but not pretty . !
 [/INST] We also know a well-known indentify as follow for $\theta= \dfrac{\pi}{7}$
$$1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta= 0$$
On the other hand, we have $7- (\!4\sin 2\theta- \tan \theta\!)^{2}= (\!\sec \theta\!)^{2}(\!1+ 2\cos 2\theta+ 2\cos 4\theta+ 2\cos 6\theta\!)= 0$
Furthermore $4\sin 2\theta- \tan \theta= \tan \theta (4\cos 2\theta+ 3)> 0\,\therefore\,4\sin 2\theta- \tan \theta= \sqrt{7}$ for $\theta= \dfrac{\pi}{7}$ (OP) .
q.e.d
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