Given $\tan 3x=4$, find the value of $\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$

Question

Given $\tan 3x=4$, find the value of $S=\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$

I expanded each of $\tan (120+x)$ and $\tan (60+x)$ getting as

$$S=\tan^2 x+\left(\frac{\tan 120+\tan x }{1-\tan 120 \tan x}\right)^2+\left(\frac{\tan 60+\tan x }{1-\tan 60 \tan x}\right)^2$$ so

$$S=\tan^2 x+\left(\frac{-\sqrt{3}+\tan x }{1+\sqrt{3} \tan x}\right)^2+\left(\frac{\sqrt{3}+\tan x }{1-\sqrt{3} \tan x}\right)^2$$

so

$$S=\tan^2 x+\frac{\left(4\tan x-\sqrt{3}(1+\tan^2x)\right)^2+\left(4\tan x+\sqrt{3}(1+\tan^2x)\right)^2}{(1-3\tan^2 x)^2}$$ so

$$S=\tan^2 x+\frac{\left(32 \tan^2 x+6(1+\tan^2 x)^2\right)}{(1-3\tan^2 x)^2}$$

but if i proceed further i dot think i will get an expression in terms of $\tan 3x$.

Answer 1

Hint:

Observe the following :

$1)$ $3\tan(3x)=\tan(x)+\tan({x+60^{\circ}})+\tan({x+120^{\circ}})$

$2)$ $\tan{60^{\circ}=\tan[(x+60^{\circ})-60^{\circ}}]$ and similar results

$3)$ $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$

$4)$ $\tan(60^{\circ})=\sqrt{3}$ and $\tan(120^{\circ})=-\sqrt{3}$

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From $(1)$, we get

$$\tan^2(x)+\tan^2({x+60^{\circ}})+\tan^2({x+120^{\circ}})=[\tan^2(3x)-2[\tan(x)\tan({x+60^{\circ}})+\tan({x+120^{\circ}})\tan({x+60^{\circ}})+\tan(x)\tan({x+120^{\circ}})]$$

From $(2),(3)$ and $(4)$ deduce that

$$\tan(x)\tan(x+60^{\circ})=\dfrac{\tan(x+60^{\circ})-\tan(x)-\sqrt{3}}{\sqrt{3}}$$ and similar results

You can surely take it from here.

Answer 2

As $a=\tan x,$

$b=\tan(60^\circ+x),\tan3(60^\circ+x)=\cdots=\tan3x,$

and $c=\tan(120^\circ+x),\tan3(120^\circ+x)=\cdots=\tan3x$

The roots of $$4=\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$$

$$\iff\tan^3y-12\tan^2y-3\tan y+4=0$$ are $a,b,c$

We need $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$