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.

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?

(1) Note first that $$\tan x\tan(2x)=\frac{\sin x\sin(2x)}{\cos x\cos(2x)}=\frac{2\sin^2x}{\cos(2x)}=\frac{1}{\cos(2x)}-1 $$ (2) It follows that $$\eqalign{ S~&\buildrel{\rm def}\over{=}~\tan\alpha\tan{2\alpha}+\tan2\alpha\tan{4\alpha}+\tan4\alpha\tan{\alpha}\cr &=-3+\frac{1}{\cos\alpha}+\frac{1}{\cos2\alpha}+\frac{1}{\cos4\alpha}\cr &=-3+\frac{\cos\alpha\cos2\alpha+\cos2\alpha\cos4\alpha+\cos4\alpha\cos\alpha}{\cos\alpha\cos2\alpha\cos4\alpha}\cr &=-3+\frac{\cos\alpha+\cos3\alpha+\cos6\alpha+\cos2\alpha+\cos5\alpha+\cos 4\alpha}{2\cos\alpha\cos2\alpha\cos4\alpha}\tag{1}\cr } $$ (3) If $\xi=e^{i\alpha}$ then we have $$1+\xi+\xi^2+\xi^3+\xi^4+\xi^5+\xi^6=0$$ Taking real parts we get $$1+\cos\alpha+\cos2\alpha+\cos3\alpha+\cos4\alpha+\cos5\alpha+\cos6\alpha=0\tag{2}$$ also $$8\cos\alpha\cos2\alpha\cos4\alpha=\frac{8\sin\alpha\cos\alpha\cos2\alpha\cos4\alpha}{\sin\alpha}=\frac{\sin8\alpha}{\sin\alpha}=1\tag{3}$$ (4) Replacing $(2)$ and $(3)$ in $(1)$ we obtain $$ S=-3+\frac{-4}{1}=-7. $$ which is the desired conclusion.