Question

The solution of the equation $\forall \theta ϵ(0,\pi )$
$\tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta$ is given by

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{5\pi }{6}$

Answer 1

Answer: (A), (B)

$\frac{\pi }{3}$
$\frac{2\pi }{3}$

$\tan \theta +\tan 2\theta =-\tan 3\theta (1-\tan \theta .\tan 2\theta )$
$\Rightarrow \frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=-\tan 3\theta$
$\tan 3\theta =\tan (n\pi -3\theta )$
$\Rightarrow \theta =\frac{n\pi }{6},nϵN$
For$(0,\pi )$
$\therefore \theta =\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{2},\frac{2\pi }{3},\frac{5\pi }{6}$
From these values $\tan \theta$ is not defined for $\frac{\pi }{2}$
and $\tan 3\theta$ is not defined for $\frac{\pi }{6},\frac{5\pi }{6}$
So, the solutions are $\frac{\pi }{3}$ and $\frac{2\pi }{3}$