Question

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

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

Answer 1

Answer: (D)

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