Question

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

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

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