Question

The solution of $\tan \left(2\theta \right)\tan \left(\theta \right)=1$ is

• A. $\frac{\mathrm{\pi }}{3}$
• B. $(6n\pm 1)\frac{\mathrm{\pi }}{6}$
• C. $(4n\pm 1)\frac{\mathrm{\pi }}{6}$
• D. $(2n\pm 1)\frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is B

$(6n\pm 1)\frac{\mathrm{\pi }}{6}$

Explanation for the correct option

Given: $\tan \left(2\theta \right)\tan \left(\theta \right)=1$

$$\begin{gathered} \Rightarrow \frac{2{\tan }^2\left(\theta \right)}{1-{\tan }^2\left(\theta \right)}=1 \\ \Rightarrow 2{\tan }^2\left(\theta \right)=1-{\tan }^2\left(\theta \right) \\ \Rightarrow 3{\tan }^2\left(\theta \right)=1 \\ \Rightarrow {\tan }^2\left(\theta \right)=\frac{1}{3} \\ \Rightarrow \tan \left(\theta \right)=\pm \frac{1}{\sqrt{3}} \\ \Rightarrow \theta =\pm {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \Rightarrow \theta =n\pi \pm \frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, option B is correct.