Since nobody has posted a solution yet I will do this now (but probably this one does not correspond well to the attribute "nice"). Let $x=e^{\frac{i\pi}{7}}$. Then $x^7=-1$. Now, by well-known identities $\sin \frac{\pi}{7}=\frac{x-x^{-1}}{2i}=\frac{x^2-1}{2ix}$ and $\tan \frac{3\pi}{7}=\frac{x^3-x^{-3}}{i(x^3+x^{-3})}=\frac{x^6-1}{i(x^6+1)}$. So we are left to prove $\frac{x^6-1}{i(x^6+1)}-4\frac{x^2-1}{2ix}=\sqrt{7}$ which is equivalent to $x(x^6-1)-2(x^2-1)(x^6+1)=i\sqrt{7}x(x^6+1)$ $\Leftrightarrow 2x^6-2x^2+x+1=i\sqrt{7}(x-1)$ $\Leftrightarrow (2x^6-2x^2+x+1)^2=-7(x-1)^2$ $\Leftrightarrow 4x^6-4x^5+4x^4-4x^3+4x^2-4x+4=0$ which is true because $x^7+1=0$ and $x+1 \ne 0$ imply $x^6-x^5+x^4-x^3+x^2-x+1=0$. Hence the result.

Tintarn wrote: $\Leftrightarrow 2x^6-2x^2+x+1=i\sqrt{7}(x-1)$ $\Leftrightarrow (2x^6-2x^2+x+1)^2=-7(x-1)^2$ This is the right way to solve this problem but there is something to develop more here because there is no equivalence between the two lines above (only one implication). To be clear, we know that $x^6-x^5+x^4-x^3+x^2-x+1=0$ which implies $(2x^6-2x^2+x+1)^2=-7(x-1)^2$ but why $2x^6-2x^2+x+1=i\sqrt{7}(x-1)$ and not $2x^6-2x^2+x+1=-i\sqrt{7}(x-1)$?

Okay, thanks for your remark, tchebytchev! In fact it's easy to see that my post above establishes that $|\tan (\frac{3\pi}{7})-4\sin (\frac{\pi}{7})|= \sqrt{7}$ so it suffices to prove that $LHS>0$. This can be easily done by rough estimations since the $\tan x$ is increasing and $\sin x$ is decreasing for $x \in (0,\frac{\pi}{2})$ This implies $LHS>\tan \frac{5\pi}{12}-4\sin \frac{\pi}{6}=2+\sqrt{3}-4 \cdot \frac{1}{2}=\sqrt{3}>0$ and hence the identity holds.

It is correct now, just there is a typo, $\tan x$ and $\sin x$ are both increasing but there is no impact of this typo to the proof because you use the correct sentence implicitly in your reasoning after. Tintarn wrote: This can be easily done by rough estimations since the $\tan x$ is increasing and $\sin x$ is decreasing for $x \in (0,\frac{\pi}{2})$