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If $$\tan(x) +\tan(x+ \frac{\pi}{3}) +\tan(x+\frac{2\pi}{3}) = 3$$ prove that $$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=1$$

hints will be appreciated

thanks.

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    What have you tried so far? Try writing the 2nd and the 3rd $tan$ as $\sin/\cos$ and adding them together, the numerator should be of the form $\sin(x+y)$ – 2017-01-31

5 Answers 5

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We try to simplify $\tan (x + \frac {\pi}{3}) + \tan (x +\frac {2\pi}{3}) $. We have $$\tan (x +\frac {\pi}{3}) + \tan (x +\frac {2\pi}{3}) $$ $$= \frac {\tan x +\sqrt {3}}{1-\sqrt {3}\tan x} + \frac {\tan x -\sqrt {3}}{1+\sqrt {3}\tan x}$$ $$= \frac {\tan x +\sqrt {3} + \sqrt {3}\tan^2 x + 3\tan x +\tan x- \sqrt {3} -\sqrt {3}\tan^2 x +3\tan x}{1-3\tan^2 x} $$ $$=\frac {8\tan x}{1-3\tan^2 x} $$

Now add this to $\tan x $ and the result follows. Hope it helps.

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    This is faster than others. – 2017-01-31
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Hint: Use these

(1) $$\tan(x+\frac{2\pi}{3}) = \tan(x- \frac{\pi}{3})$$ (2) $$\tan x+\tan y=\frac{\sin(x+y)}{\cos x\cos y}$$

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    your first hint is not clear, how do you get this – 2017-01-31
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    @TaylorTed Substitute it in equation and use trigonometric relations. – 2017-01-31
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    i mean proof for first hint – 2017-01-31
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    Proof $$\tan(x+\frac{2\pi}{3}) = \tan(x+\pi- \frac{\pi}{3})= \tan(\pi+x- \frac{\pi}{3})=\frac{\tan\pi+\tan(x-\frac{\pi}{3})}{1-\tan\pi\tan(x-\frac{\pi}{3})}=\tan(x-\frac{\pi}{3})$$ – 2017-01-31
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[ tan(x) + √3 ] .......[ tan(x) - √3 ] -------------------- + --------------------- = 3 - tan (x) [1 - √3 tan(x)]........[1 + √3 tan(x)]

tan(x) + √3 tan²(x) + √3 + 3 tan(x) + tan(x) - √3 tan²(x) - √3 + 3 tan(x) --------------------------------------... = 3 - tan (x) LCM = [1 - √3 tan(x)]*[1 + √3 tan(x)]

8 tan(x) -------------------- = 3 - tan (x) ; use (a + b)*)(a - b) = a² - b² in the DEN [1 - 3 tan²(x)]

8 tan(x) = 3 - 3*3 tan²(x) - tan(x) + 3 tan³(x) ; cross multiply

3 tan³(x) - 9 tan²(x) - 9 tan (x) + 3 = 0 ; or

tan³(x) - 3 tan²(x) - 3 tan (x) + 1 = 0 -----> Proved

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    This could really use some LaTeX/MathJax formatting! – 2017-01-31
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    new users are forgiven – 2017-01-31
1

Let (for brevity) $y=\tan x$ and note that $$\tan \left(x+\frac\pi3\right)=\frac{y+\sqrt 3}{1-y\sqrt 3}$$ $$\tan \left(x+\frac{2\pi}3\right)=\frac{y-\sqrt 3}{1+y\sqrt 3}$$ Now we have that $$y+\frac{y+\sqrt 3}{1-y\sqrt 3}+\frac{y-\sqrt 3}{1+y\sqrt 3}=3$$

Simplify the LHS and then divide by $3$.

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If $\tan3x=\tan 3y$

$$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3y$$

$$\implies \tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$

Again, as $\tan3x=\tan 3y$

$3x=180^\circ n+3y$ where $n$ is any integer $\implies x=60^\circ n+y$ where $n\equiv0,1,2\pmod3$

So, the corresponding roots are $$\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)$$

Now apply Vieta's formula.