If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$. 
If $\alpha =2\pi/7$, find $\tan\alpha \tan 2\alpha + \tan 2\alpha \tan 4\alpha + \tan 4\alpha \tan \alpha$.

I tried by $7\alpha = 2\pi$
$4\alpha =2\pi-3\alpha$
$\sin 4\alpha = \sin 3\alpha$
But couldn't reach solution Please help
 A: Since $$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8},$$
we obtain
$$\tan\frac{2\pi}{7}\tan\frac{4\pi}{7}+\tan\frac{2\pi}{7}\tan\frac{8\pi}{7}+\tan\frac{4\pi}{7}\tan\frac{8\pi}{7}=$$
$$=8\left(\cos\frac{8\pi}{7}\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\sin\frac{8\pi}{7}+\cos\frac{2\pi}{7}\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}\right)=$$
$$=4\left(\cos\frac{8\pi}{7}\left(\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}\right)+\cos\frac{4\pi}{7}\left(\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}\right)+\cos\frac{2\pi}{7}\left(\cos\frac{4\pi}{7}-\cos\frac{2\pi}{7}\right)\right)=$$
$$=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{2\pi}{7}-\cos\frac{14\pi}{7}\right)+$$
$$+2\left(\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}-1\right)+$$
$$+2\left(\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}-1\right)=$$
$$=2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}-3\right)=$$
$$=2\left(\frac{2\sin\frac{\pi}{7}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)}{2\sin\frac{\pi}{7}}-3\right)=$$
$$=2\left(\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}-3\right)=2\left(-\frac{1}{2}-3\right)=-7.$$
Done!
A: $\displaystyle \tan \frac{k \pi}{7}; 1 \le k \le 6$ are roots of 
$\tan 7 \theta = 0$ which gives the polynomial $$y^6 - \binom{7}{2} y^4 + \binom{7}{4} y^2- 7=0$$ with the substitution $ y = \tan \theta$
This also yields that $\displaystyle \tan^2 \frac{k \pi}{7}; k={2,4,8}$ are roots to the polynomial $$y^3 - \binom{7}{2} y^2 + \binom{7}{4} y - 7=0$$ so that from Vieta's formula $$\tan^2 \frac{2 \pi}{7}+\tan^2 \frac{4 \pi}{7}+\tan^2 \frac{8 \pi}{7} = 21$$
Further we have $$\tan \frac{2 \pi}{7}+\tan \frac{4 \pi}{7}+\tan \frac{8 \pi}{7} =  \tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7}$$
But from Vieta, $$tan^2 \frac{2 \pi}{7} \tan^2 \frac{4 \pi}{7} \tan^2 \frac{8 \pi}{7} = 7$$ and hence $$\tan \frac{2 \pi}{7}\tan \frac{4 \pi}{7}\tan \frac{8 \pi}{7} = -\sqrt 7$$
Using the identity 2$$\sum ab =(a+b+c)^2 -(a^2+b^2+c^2)$$ we obtain the required sum as $-7$
