Question

If $\tan 2\theta \tan \theta =1$, then the general value of $\theta$ is :

• A. $(n+\frac{1}{2})\frac{\pi }{3}$
• B. $(n+\frac{1}{2})\pi$
• C. $(2n\pm \frac{1}{2})\frac{\pi }{2}$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

$\begin{array}{c}\tan 2\theta \cdot \tan \theta =1\\ \Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }-{\tan }^2\theta =1\left[\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }\right]\\ \Rightarrow 2{\tan }^2\theta =1\\ \Rightarrow {\tan }^2\theta =\frac{1}{3},\tan \theta =\frac{1}{\sqrt{3}},\theta =\left(n\pi +\frac{\pi }{6}\right)\\ nEI\end{array}$