Question

In $(0,\pi )$, the number of solutions of the equation $\tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta$ is

• A. $7$
• B. $5$
• C. $4$
• D. $2$

Answer 1

Answer: (B)

$5$

$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-(\tan A\tan B+\tan B\tan C+\tan C\tan A)}$

given $A=\theta ,B=2\theta ,C=3\theta$

from given equation

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

so,

$\tan (\theta +2\theta +3\theta )=0$

$tan6\theta =0$

$\Rightarrow \theta =\frac{n\pi }{6},n\in I$

$n=1,\frac{\pi }{6}n=2,\frac{\pi }{3}n=3,\frac{\pi }{2}n=4,\frac{2\pi }{3}n=5,\frac{5\pi }{6}n=6,\pi$ $(0,\pi )$ so 5 answer