Question

If $\tan x+\tan \left( x+\dfrac{\pi }{3} \right)+\tan \left( x+\dfrac{2\pi }{3} \right)=3$, then prove that \[\dfrac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}=1\].

Answer 1

Hint: We will be using the concept of trigonometric identities to solve the problem. We will first expand the given equation with the help of trigonometric formula that,
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ to further simplify the problem and then simplify the solution.

Complete step-by-step answer:

Now, we have been given that $\tan x+\tan \left( x+\dfrac{\pi }{3} \right)+\tan \left( x+\dfrac{2\pi }{3} \right)=3$.

Now, we will be using the trigonometric identity that,

$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$

Therefore, we have,

$\tan x+\dfrac{\tan x+\tan \left( \dfrac{\pi }{3} \right)}{1-\tan x\tan \left( \dfrac{\pi }{3}

\right)}+\dfrac{\tan \left( x \right)+\tan \left( \dfrac{2\pi }{3} \right)}{1-\tan \left( x

\right)\tan \left( \dfrac{2\pi }{3} \right)}=3$

Now, we will be using the fact that,

$\begin{align}

  & \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3} \\

 & \tan \left( \dfrac{2\pi }{3} \right)=-\sqrt{3} \\

\end{align}$

So, using this we have,

$\tan x+\dfrac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}+\dfrac{\tan \left( x \right)-\sqrt{3}}{1+\sqrt{3}\tan \left( x \right)}=3$

Now, we will take $\left( 1-3{{\tan }^{2}}x \right)$ as LCM. So, we have,

$\dfrac{\tan x\left( 1-3{{\tan }^{2}}x \right)+\left( \tan x+\sqrt{3} \right)\left( 1+\sqrt{3}\tan x

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

Now, on simplifying the numerator by further expanding the terms we have,

$\begin{align}

  & \dfrac{\tan x-3{{\tan }^{3}}x+\tan x+\sqrt{3}{{\tan }^{2}}x+\sqrt{3}+3\tan x+\tan x-\sqrt{3}{{\tan }^{2}}x-\sqrt{3}+3\tan x}{1-3{{\tan }^{2}}x}=3 \\

 & \dfrac{9\tan x-3{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}=3 \\

\end{align}$

Now, we will take 3 common numerators. So, we have,

$\begin{align}

  & \dfrac{3\left( 3\tan x-{{\tan }^{3}}x \right)}{1-3{{\tan }^{2}}x}=3 \\

 & \dfrac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}=1 \\

\end{align}$

So, we have proved that,

\[\dfrac{3\tan x-{{\tan }^{3}}x}{1-3{{\tan }^{2}}x}=1\] if $\tan x+\tan \left( x+\dfrac{\pi }{3}

\right)+\tan \left( x+\dfrac{2\pi }{3} \right)=3$.


Note: It is important to note that we have used the trigonometric formula that $\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$ to simplify the data given to us i.e. $\tan x+\tan \left( x+\dfrac{\pi }{3} \right)+\tan \left( x+\dfrac{2\pi }{3} \right)=3$ and then simplify the given equation further to complete the proof.