Question

The general value of $\theta$ satisfying equation $tan3\theta -tan2\theta -tan\theta =0$

• A. $n\pi$
• B. $\frac{n\pi }{4}$
• C. $\frac{n\pi }{2}$
• D. $\frac{n\pi }{3}$

Answer 1

The correct option is D $\frac{n\pi }{3}$

As $2\theta +\theta =3\theta \Rightarrow \tan (2\theta +\theta )=\tan 3\theta$

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

$\Rightarrow \tan 2\theta +\tan \theta =\tan 3\theta -\tan \theta \tan 2\theta \tan 3\theta \Rightarrow \tan 3\theta -\tan 2\theta -\tan \theta =\tan \theta \tan 2\theta \tan 3\theta$

Now
$\therefore \tan 3\theta -\tan 2\theta -\tan \theta =0\Rightarrow \tan \theta \tan 2\theta \tan 3\theta =0$

$\Rightarrow \theta =n\pi$ or $2\theta =n\pi$ or $3\theta =n\pi$

$\Rightarrow \theta =\frac{n\pi }{3}$