3

Let's assume that $\tan(x) = y$.

So, $\tan\Big(x+\dfrac{\pi}{3}\Big) = \dfrac{\tan(x) + \tan\Big(\dfrac{\pi}{3}\Big)}{1-\tan(x)\tan\Big(\dfrac{\pi}{3}\Big)} = \dfrac{y+\sqrt{3}}{1-\sqrt{3}y}$

Similarly, $\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{y-\sqrt{3}}{1+\sqrt{3}y}$

Also, $\tan\Big(\dfrac{\pi}{3}-x\Big) = -\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{\sqrt{3}-y}{1+\sqrt{3}y}$

Now, $\tan(x)\tan\Big(x+\dfrac{\pi}{3}\Big)+\tan(x)\tan\Big(\dfrac{\pi}{3}-x\Big)+\tan\Big(x+\dfrac{\pi}{3}\Big)\tan\Big(x-\dfrac{\pi}{3}\Big)$ $$ = y\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)+y\Big(\dfrac{\sqrt{3}-y}{1+\sqrt{3}y}\Big)+\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)\Big(\dfrac{y-\sqrt{3}}{1+\sqrt{3}y}\Big)$$ $$ = y\Big(\dfrac{(y+\sqrt{3})(1+\sqrt{3}y)+(1-\sqrt{3}y)(\sqrt{3}-y)}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$ $$ = y\Big(\dfrac{y+\sqrt{3}y^2+\sqrt{3}+3y+\sqrt{3}-y-3y+\sqrt{3}y^2}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$ $$ = \dfrac{2\sqrt{3}y+2\sqrt{3}y^3+y^2-3}{1-3y^2}$$

This is how much I've been able to simplify the expression but I'm unable to continue. I am familiar with the values of trigonometric functions at multiples and sub multiples of angles and I think the solution would involve their use (as the question has been taken from that very chapter).

Thanks!

Rajdeep Sindhu
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  • This identity is clearly not true. Using $x=2\pi/3$ gives$$3=-3$$ – Peter Foreman Aug 21 '20 at 21:52
  • If you graph the LHS as a function of $x$, you will see that it is not constant. That would have saved you a lot of effort. – quasi Aug 21 '20 at 22:14
  • One of us must have missed something. Taking $x=2\pi/3$ gives $\tan(2\pi/3)\tan(\pi/3)=-3$, and the graph of the left hand side looks to be constant. Mathematica also verified the equality, for what that's worth. – pancini Aug 21 '20 at 22:28
  • But by @Elliot G's (currently deleted) answer, the identity would be true if you replace $\tan(\pi/3-x)$ by $\tan(x-\pi/3)$. – quasi Aug 21 '20 at 22:29
  • Ahhh, that's what I missed. – pancini Aug 21 '20 at 22:29
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    @Elliot G: You should undelete it and indicate that you are solving what was undoubtedly the intended problem. – quasi Aug 21 '20 at 22:31

2 Answers2

3

As written, this statement is false, as per @quasi 's comment. I suspect the identity should have used $\tan(x-\pi/3)$ rather than $\tan(\pi/3-x)$. In this case, by the angle-sum formula for tangent,

$$\tan(x)\tan\left(x+\frac\pi3\right)=\tan(x)\left(\frac{\tan x+\sqrt 3}{1-\sqrt 3\tan x}\right)=\tan(x)\left(\frac{4\tan x+\sqrt 3\tan^2 x+\sqrt 3}{1-3\tan^2 x}\right)$$ and

$$\tan(x)\tan\left(x-\frac\pi3\right)=\tan(x)\left(\frac{\tan x-\sqrt 3}{1+\sqrt 3\tan x}\right)=\tan(x)\left(\frac{4\tan x-\sqrt 3\tan^2 x-\sqrt 3}{1-3\tan^2 x}\right)$$

and

$$\tan\left(x+\frac\pi3\right)\tan\left(x-\frac\pi3\right)=\left(\frac{\tan x+\sqrt 3}{1-\sqrt 3\tan x}\right)\left(\frac{\tan x-\sqrt 3}{1+\sqrt 3\tan x}\right)=\frac{\tan^2x-3}{1-3\tan^2x}.$$

The sum is then

$$\frac{\tan(x)(8\tan x)+\tan^2 x-3}{1-3\tan^2 x}=-3.$$

pancini
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0

Like Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$

$$t_0=\tan y,t_1=\tan\left(y-\dfrac\pi3\right),t_2=\tan\left(y+\dfrac\pi3\right)$$ are the roots of

$$t^3-(3\tan3y)t^2-3t+3\tan3y=0$$

Using Vieta's formula $$t_0t_1+t_1t_2+t_2t_0=\dfrac{-3}1$$

$$t_0+t_1+t_2=\dfrac{3\tan3y}1$$

$$t_0t_1t_2=\dfrac{-3\tan3y}1$$

if $t_0,t_1,t_2$ are finite i.e., $y+\dfrac{n\pi}3\ne m\pi+\dfrac\pi2; n=-1,0,1$

lab bhattacharjee
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