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 . !
 A: It's enough to prove $\big(4\sin(2\pi/7)-\tan(\pi/7)\big)^2=7$. But if $\zeta=\exp(2\pi i/7)$ then
$$4\sin\frac{2\pi}{7}-\tan\frac{\pi}{7}=\frac{1}{i}\Big(2\zeta-\frac{2}{\zeta}-\frac{\zeta-1}{\zeta+1}\Big)=\frac{2\zeta^3+\zeta^2-\zeta-2}{i\zeta(\zeta+1)}$$
and $(2\zeta^3+\zeta^2-\zeta-2)^2+7\zeta^2(\zeta+1)^2=4(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)=0$.
A: Hint:
Rewrite this as
$\tan\frac{\pi}{7}(8\cos^2\frac\pi7 -1)=\sqrt7$
$\tan^2\frac{\pi}{7}(8\cos^2\frac\pi7 -1)^2=7$
$(1-\cos^2\frac{\pi}{7})(8\cos^2\frac\pi7 -1)^2=7\cos^2\frac\pi7$
$(1-\cos\frac{2\pi}{7})(4\cos\frac{2\pi}{7}+3)^2=7\cos\frac{2\pi}{7}+7$
$8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=0$
All we need is to prove that $\cos\frac{2\pi}7$ is a root of $8x^3+4x^2-4x-1=0$
Start from $\cos3(\frac{2\pi}{7})=\cos4(\frac{2\pi}{7})$ ...
A: If $\tan\dfrac\pi7=t$
we need to show $\sqrt7=\dfrac{8t}{1+t^2}-t$
$t^3+\sqrt7t^2-7t+\sqrt7=0$
Now use my answer from If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$
A: 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
