MHB Another trigonometric equality

[Albert1]

the units of all angles :degree
gien :$tan\, x\,+tan\,(x+60)\,-\,tan(60-x)=3tan(3x)$
prove :$tan^2\, x\,+tan^2\,(x+60)\,+\,tan^2(60-x)=9tan^2(3x)+6$

 

[anemone]

My solution:

Spoiler

First notice that

i.
$\begin{align*}\small\tan 3x&=\dfrac{\tan x(3-\tan^2 x)}{1-3\tan^2x}\\&= \tan x \left(\dfrac{(\sqrt{3})^2-\tan^2 x}{1^2-(\sqrt{3} \tan x)^2} \right)\\&=\tan x \left(\dfrac{(\sqrt{3}-\tan x)(\sqrt{3}+\tan x)}{(1+\sqrt{3} \tan x)(1-\sqrt{3} \tan x)} \right)\\&=\tan x \tan(60-x) \tan (60+x)\end{align*}$

ii.
$\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2x}\rightarrow\,\,\tan^3 x+\tan3x=3\tan x+3\tan^2 x \tan 3x$We are given $\tan x^{\circ}+\tan(60+x)^{\circ}-\tan(60-x)^{\circ}=3\tan 3x^{\circ}$

If we rewrite it as $\tan(60+x)^{\circ}-\tan(60-x)^{\circ}=3\tan 3x^{\circ}-\tan x^{\circ}$ and squaring it, we get:

$\tan^2(60+x)^{\circ}+\tan^2(60-x)^{\circ}-2\tan(60+x)^{\circ}\tan(60-x)^{\circ}=9\tan^2 3x^{\circ}+\tan^2 x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}$

Modifying it a bit we get

$\begin{align*}\tan^2 x^{\circ}+\tan^2(60+x)^{\circ}+\tan^2(60-x)^{\circ}&=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+2\tan(60+x)^{\circ}\tan(60-x)^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+\dfrac{2\tan 3x^{\circ}}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+\dfrac{2(\tan^3 x^{\circ}+\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+\dfrac{2(3\tan x^{\circ}+3\tan^2 x^{\circ}\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+6+6\tan 3x^{\circ}\tan x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+6\end{align*}$

 

[kaliprasad]

My solution:

Spoiler

First notice that

i. $\small\tan 3x=\dfrac{\tan x(3-\tan^2 x)}{1-3\tan^2x}= \tan x \left(\dfrac{(\sqrt{3})^2-\tan^2 x}{1^2-(\sqrt{3} \tan x)^2} \right)=\tan x \left(\dfrac{(\sqrt{3}-\tan x)(\sqrt{3}+\tan x)}{(1+\sqrt{3} \tan x)(1-\sqrt{3} \tan x)} \right)=\tan x \tan(60-x) \tan (60+x)$

ii. $\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2x}\rightarrow\,\,\tan^3 x+\tan3x=3\tan x+3\tan^2 x \tan 3x$We are given $\tan x^{\circ}+\tan(60+x)^{\circ}-\tan(60+x)^{\circ}=3\tan 3x^{\circ}$

If we rewrite it as $\tan(60+x)^{\circ}-\tan(60+x)^{\circ}=3\tan 3x^{\circ}-\tan x^{\circ}$ and squaring it, we get:

$\tan^2(60+x)^{\circ}+\tan^2(60+x)^{\circ}-2\tan(60+x)^{\circ}\tan(60+x)^{\circ}=9\tan^2 3x^{\circ}+\tan^2 x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}$

Modifying it a bit we get

$\begin{align*}\tan^2 x^{\circ}+\tan^2(60+x)^{\circ}+\tan^2(60+x)^{\circ}&=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+2\tan(60+x)^{\circ}\tan(60+x)^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+2\tan^2 x^{\circ}+\dfrac{2\tan 3x^{\circ}}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+\dfrac{2(\tan^3 x^{\circ}+\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+\dfrac{2(3\tan x^{\circ}+3\tan^2 x^{\circ}\tan 3x^{\circ})}{\tan x^{\circ}}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+6+6\tan 3x^{\circ}\tan x^{\circ}-6\tan 3x^{\circ}\tan x^{\circ}\\&=9\tan^2 3x^{\circ}+6\end{align*}$

the 3rd term in a couple of lines should be $\tan\left({60-x}\right)$

 

[anemone]

the 3rd term in a couple of lines should be $\tan\left({60-x}\right)$

Thanks, kali! I will fix it right away!