Question

The solution of $\tan 2\theta \tan \theta =1$ is

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

Answer 1

Answer: (B)

$\tan 2\theta \tan \theta =1$
$\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }\tan \theta =1$
$\Rightarrow 2{\tan }^2\theta =1-{\tan }^2\theta$
$\Rightarrow 3{\tan }^2\theta =1$
$\Rightarrow \tan \theta =\pm \frac{1}{\sqrt{3}}=\tan \left(\pm \frac{\pi }{6}\right)$
$\Rightarrow \theta =n\pi \pm \frac{pi}{6}(m\in Z)$
= $(6n\pm 1)\frac{\pi }{6}$