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

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

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.
