<s>[INST] If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha.$ If $\alpha = \frac{2\pi}{7}$ then the find the value of $\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$ 
My 1st  approach : 
$\tan(\alpha +2\alpha +4\alpha) = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $
$\Rightarrow 0 = \frac{\tan\alpha +\tan2\alpha +\tan4\alpha -\tan\alpha \tan2\alpha -\tan2\alpha \tan4\alpha -\tan4\alpha \tan\alpha}{1-(\tan\alpha \tan2\alpha +\tan2\alpha \tan4\alpha +\tan\alpha \tan4\alpha)} $ which doesn't give me any solution. 
My IInd approach : 
U\sing Euler substitution : 
\since $\cos\theta +i\sin\theta = e^{i\theta} $.....(i) and $\cos\theta -i\sin\theta =e^{-i\sin\theta}$....(ii)
Adding (i) and (ii) we get $\cos\theta =\frac{e^{i\theta} +e^{-i\theta}}{2}$  and subtracting (i) and (ii) we get $\sin\theta =\frac{e^{i\theta} -e^{-i\theta}}{2}$ 
By u\sing this we can write : $$\tan\alpha .\tan2\alpha +\tan2\alpha \tan4\alpha +\tan4\alpha \tan\alpha$$ as $$\frac{1}{4}\left[ (e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}) (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}}) + (e^{\frac{i4\pi}{7}} -e^{\frac{-i4\pi}{7}})(e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) + (e^{\frac{i8\pi}{7}} -e^{\frac{-i8\pi}{7}}) (e^{\frac{i\pi}{7}} -e^{\frac{-i\pi}{7}})\right]$$
$$\large= e^{i\frac{6\pi}{7}}-e^{\frac{i2\pi}{7}}-e^{\frac{-i2\pi}{7}} +e^{\frac{-i6\pi}{7}} +e^{\frac{i3\pi}{7}}-e^{\frac{-i5\pi}{7}}-e^{\frac{i5\pi}{7}} +e^{\frac{-3\pi}{7}} +e^0 -e^{\frac{i2\pi}{7}} -e^{\frac{-i2\pi}{7}}+e^0$$
Can anybody please suggest whether this is my correct approach or not. please guide further... Thanks.
 [/INST] Not using Euler's formula which I don't think the best way for this
Let  $\displaystyle a=\tan A,b=\tan2A,c=\tan4A$ where $A+2A+4A=n\pi$ where $7\nmid n$
As $\displaystyle\tan(n\pi-rA)=-\tan rA,\tan6A=-\tan A=-a$ etc.
Using  Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$,
$\displaystyle a+b+c=abc$
Now using Sum of tangent functions where arguments are in specific arithmetic series,
$\displaystyle\tan7x=\frac{\binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x}{1-\binom72\tan^2x+\binom74\tan^4x-\binom76\tan^6x}$
If $\displaystyle\tan7A=0,7A=m\pi$ where $m$ is any integer
$\displaystyle\implies A=\frac{m\pi}7$ where $0\le m\le6$
So, $\pm a,\pm b,\pm c,\tan0=0$ are the roots of 
$\displaystyle \binom71\tan x-\binom73\tan^3x+\binom75\tan^5x-\tan^7x=0$
$\displaystyle\iff\tan^7x-21\tan^5x+35\tan^3x-7\tan x=0$
So, $\pm a,\pm b,\pm c$ are the roots of 
$\displaystyle \tan^6x-21\tan^4x+35\tan^2x-7=0\ \ \ \  (1)$
Now the equation whose roots are $\pm a,\pm b,\pm c$ is 
$\displaystyle(y-a)(y-b)(y-c)(y+a)(y+b)(y+c)=0$
$\displaystyle\iff(y^2-a^2)(y^2-b^2)(y^2-c^2)=0$
$\displaystyle\iff y^6-(a^2+b^2+c^2)y^4+(a^2b^2+b^2c^2+c^2a^2)y^2-a^2b^2c^2=0\ \ \ \ (2)$
If we write $\displaystyle a+b+c=abc=S$  and  $\displaystyle ab+bc+ca=T,$
$\displaystyle a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=S^2-2T$
and $\displaystyle a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=T^2-2S^2$
So, $(2)$ becomes $\displaystyle y^6-(S^2-2T)y^4+(T^2-2S^2)y^2-S^2=0$
Comparing with $\displaystyle(1), S^2=7,S^2-2T=21, T^2-2S^2=35$
Can you find the required $T$ from here?
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