Question

If $\tan \left(\theta \right)+\tan \left(\theta +\frac{\mathrm{\pi }}{3}\right)+\tan \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=3$, then which of the following is equal to $1$?

• A. $\tan \left(2\theta \right)$
• B. $\tan \left(3\theta \right)$
• C. $\tan \left(\theta \right)$
• D. ${\tan }^3\left(\theta \right)$

Answer 1

The correct option is B

$\tan \left(3\theta \right)$

Explanation for correct option

Given: $\tan \left(\theta \right)+\tan \left(\theta +\frac{\mathrm{\pi }}{3}\right)+\tan \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=3$

$$\begin{gathered} \Rightarrow \tan \left(\theta \right)+\tan \left(\theta +\frac{\mathrm{\pi }}{3}\right)+\tan \left(\theta +\frac{2\mathrm{\pi }}{3}\right)=3\because \left[\tan \left(x+y\right)=\frac{\tan \left(x\right)+\tan \left(y\right)}{1-\tan \left(x\right)\tan \left(y\right)}\right] \\ \Rightarrow \tan \left(\theta \right)+\frac{\tan \left(\theta \right)+\tan \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{1-\tan \left(\theta \right)\tan \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}+\frac{\tan \left(\theta \right)+\tan \left({\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}{1-\tan \left(\theta \right)\tan \left({\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}=3 \\ \Rightarrow \tan \left(\theta \right)+\frac{\tan \left(\theta \right)+\sqrt{3}}{1-\sqrt{3}\tan \left(\theta \right)}+\frac{\tan \left(\theta \right)-\sqrt{3}}{1+\sqrt{3}\tan \left(\theta \right)}=3\because \left[\tan \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{3}\right] \\ \Rightarrow \frac{\tan \left(\theta \right)+\sqrt{3}{\tan }^2\left(\theta \right)+3\tan \left(\theta \right)+\sqrt{3}+\tan \left(\theta \right)-\sqrt{3}{\tan }^2\left(\theta \right)+3\tan \left(\theta \right)-\sqrt{3}}{\left(1-\sqrt{3}\tan \left(\theta \right)\right)\left(1+\sqrt{3}\tan \left(\theta \right)\right)}=3-\tan \left(\theta \right) \\ \Rightarrow \tan \left(\theta \right)+\frac{8\tan \left(\theta \right)}{\left(1-{\left(\sqrt{3}\tan \left(\theta \right)\right)}^2\right)}=3 \\ \Rightarrow \frac{\tan \left(\theta \right)-3{\tan }^3\left(\theta \right)+8\tan \left(\theta \right)}{1-3{\tan }^2\left(\theta \right)}=3 \\ \Rightarrow \frac{9\tan \left(\theta \right)-3{\tan }^3\left(\theta \right)}{1-3{\tan }^2\left(\theta \right)}=3 \\ \Rightarrow 3\frac{3\tan \left(\theta \right)-{\tan }^3\left(\theta \right)}{1-3{\tan }^2\left(\theta \right)}=3 \\ \Rightarrow 3\tan \left(3\theta \right)=3\because \left[\frac{3\tan \left(x\right)-{\tan }^3\left(x\right)}{1-3{\tan }^2\left(x\right)}=\tan \left(3x\right)\right] \\ \Rightarrow \tan \left(3\theta \right)=1 \end{gathered}$$

Hence, option B is correct.