Prove tan x + tan(x+60°) + tan(x+120°)= 3 tan 3x

[Albert1]

Prove

is the statement true ?

$ tan \,x + tan(x+60^o) + tan(x+120^o)= 3\, tan\, 3x$

 

[Opalg]

Re: Prove

[sp]If $t=\tan\theta$ then $t= \tan x$, $t=\tan(x+60^\circ)$ and $t=\tan(x+120^\circ)$ are the roots of the equation $\tan 3\theta = \tan 3x$. But $\tan 3\theta = \dfrac{t^3-3t}{3t^2-1}$. So the equation can be written as $t^3-3t = (3t^2-1)\tan3x$, or $t^3 - (3\tan3x)t^2 -3t + \tan3x = 0.$ The sum of the roots of that equation is $3\tan3x.$ Hence $\tan x + \tan(x+60^\circ) + \tan(x+120^\circ) = 3\tan3x$ (with the proviso that the tan function becomes infinite at some points, in which case both sides of the equation will be undefined).[/sp]

 

[Albert1]

Re: Prove

[sp]If $t=\tan\theta$ then $t= \tan x$, $t=\tan(x+60^\circ)$ and $t=\tan(x+120^\circ)$ are the roots of the equation $\tan 3\theta = \tan 3x$. But $\tan 3\theta = \dfrac{t^3-3t}{3t^2-1}$. So the equation can be written as $t^3-3t = (3t^2-1)\tan3x$, or $t^3 - (3\tan3x)t^2 -3t + \tan3x = 0.$ The sum of the roots of that equation is $3\tan3x.$ Hence $\tan x + \tan(x+60^\circ) + \tan(x+120^\circ) = 3\tan3x$ (with the proviso that the tan function becomes infinite at some points, in which case both sides of the equation will be undefined).[/sp]

very good solution :)

 

[Albert1]

Re: Prove

is the statement true ?

$ tan \,x + tan(x+60^o) + tan(x+120^o)= 3\, tan\, 3x----(1)$

Spoiler

$ tan \,x + tan(x+60^o) + tan(x+120^o)$
$ =tan \,x + tan(60^o+x) - tan(60^o-x)$
$= tan \,x + tan\,2x[(1+tan(60^o+x)\times tan(60^o-x)]$
let :$tan\, x=t$
we have:$t+\dfrac {2t}{1-t^2}\times(1+\dfrac{3-t^2}{1-3t^2})$
$=t+\dfrac{2t}{1-t^2}(\dfrac{4-4t^2}{1-3t^2})$
$=t+\dfrac{8t}{1-3t^2}$
$=\dfrac{9t-3t^3}{1-3t^2}$
$=3\,tan\,3x---(1)$
here :$1-3t^2\neq 0$
now (1) has been proved :
please use(1) and find the value of :
$tan\,1^o+tan\,5^o+tan\,9^o+------+tan\,177^o=?$

 

[CellCoree]

To prove this statement, we will use the trigonometric identity: $tan(A+B) = \frac{tan\,A + tan\,B}{1 - tan\,A\,tan\,B}$.

Substituting $A = x$ and $B = 60^o$, we get:

$tan(x+60^o) = \frac{tan\,x + tan\,60^o}{1 - tan\,x\,tan\,60^o}$

Since $tan\,60^o = \sqrt{3}$, we can rewrite the equation as:

$tan(x+60^o) = \frac{tan\,x + \sqrt{3}}{1 - \sqrt{3}\,tan\,x}$

Using the same identity, we can also write $tan(x+120^o)$ as:

$tan(x+120^o) = \frac{tan\,x + tan\,120^o}{1 - tan\,x\,tan\,120^o}$

Since $tan\,120^o = -\sqrt{3}$, we get:

$tan(x+120^o) = \frac{tan\,x - \sqrt{3}}{1 + \sqrt{3}\,tan\,x}$

Now, substituting these values in the original equation, we get:

$tan\,x + \frac{tan\,x + \sqrt{3}}{1 - \sqrt{3}\,tan\,x} + \frac{tan\,x - \sqrt{3}}{1 + \sqrt{3}\,tan\,x} = 3\,tan\,3x$

Simplifying the left side, we get:

$tan\,x + \frac{2\,tan\,x}{1 - 3\,tan^2\,x} = 3\,tan\,3x$

Using the identity $tan(3x) = \frac{3\,tan\,x - tan^3\,x}{1 - 3\,tan^2\,x}$, we can rewrite the equation as:

$tan\,x + \frac{2\,tan\,x}{1 - 3\,tan^2\,x} = \