Question

If $\tan \left(2\theta \right)\tan \left(\theta \right)=1$, then the general value of $\theta$ is

• A. $\left(n+\frac{1}{2}\right)\frac{\pi }{3}$
• B. $\left(n+\frac{1}{2}\right)\pi$
• C. $\left(2n\pm \frac{1}{2}\right)\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

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

Use the formula $\tan \left(2A\right)=\frac{2\tan \left(A\right)}{1-{\tan }^2\left(A\right)}$

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

Hence, the correct option is (D).