Question

If $\sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1$, then the general value of $\theta$ is

• A. $n\pi +\frac{\pi }{5}$
• B. $\left(n+\frac{1}{6}\right)\frac{\pi }{5}$
• C. $\left(2n\pm \frac{1}{6}\right)\frac{\pi }{5}$
• D. $\left(n+\frac{1}{3}\right)\frac{\pi }{5}$

Answer 1

Answer: (B)

$\sqrt{3}\tan 2\theta +\sqrt{3}\tan 3\theta +\tan 2\theta \tan 3\theta =1$
$\Rightarrow \sqrt{3}(\tan 2\theta +\tan 3\theta )=1-\tan 2\theta \tan 3\theta$
$\Rightarrow \frac{\tan 2\theta +\tan 3\theta }{1-\tan 2\theta \tan 3\theta }=\frac{1}{\sqrt{3}}\Rightarrow \tan 5\theta =\tan \frac{\pi }{6}$
$\Rightarrow 5\theta =n\pi +\frac{\pi }{6}$
$\Rightarrow \theta =\left(n+\frac{1}{6}\right)\frac{\pi }{5}$