Question

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

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

Answer 1

Answer: (B)

$\left(n+\frac{1}{6}\right)\frac{\pi }{5}$

$\sqrt{3}(\tan 2\theta +\tan 3\theta )=1-\tan 2\theta \tan 3\theta$

$\Rightarrow \frac{\tan 2\theta +\tan 3\theta }{1-\tan 2\theta \tan 3\theta }=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan 5\theta =\frac{1}{\sqrt{3}}$

$\Rightarrow \tan 5\theta =\tan \pi /6$

$\Rightarrow 5\theta =n\pi +\pi /6\Rightarrow \theta =\frac{1}{5}(n\pi +\pi /6)=(n+\frac{1}{6})\pi /5$ [B]