Question

Assertion :(A) : In $(0,\pi )$ the number of solutions of the equation $tan\theta +tan2\theta +tan3\theta =tan\theta tan2\theta tan3\theta$ is two Reason: (R) :$tan6\theta$ is not defined at $\theta =(2n+1)\frac{\pi }{12},n\in l$

• A. Both (A) and (R) are individually true and (R) is the correct explanation of (A)
• B. (R) is not the correct explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

The correct option is B (R) is not the correct explanation of (A)

$3\theta =\theta +2\theta \Rightarrow \tan \left(3\theta \right)=\tan (\theta +2\theta )\Rightarrow \tan 3\theta =\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }\Rightarrow \tan 3\theta -\tan \theta \tan 2\theta \tan 3\theta =\tan \theta \tan 2\theta \Rightarrow \tan 3\theta -\tan \theta -\tan 2\theta =\tan \theta \tan 2\theta \tan 3\theta \therefore \tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta \Rightarrow \tan \theta +\tan 2\theta +\tan 3\theta =\tan 3\theta -\tan \theta -\tan 2\theta \Rightarrow \tan \theta +\tan 2\theta =0$
$\Rightarrow \frac{\sin \theta }{cos\theta }+\frac{\sin 2\theta }{cos2\theta }=0$
$\Rightarrow \frac{\sin \theta cos2\theta +\sin 2\theta cos\theta }{cos\theta cos2\theta }=0$
$\Rightarrow \sin 3\theta =0$
$\Rightarrow 3\theta =n\pi \Rightarrow \theta =\frac{n\pi }{3}$
As $\tan A$ is not define for $A=(2n+1)\frac{\pi }{2}$
$\Rightarrow \tan 6\theta$ is not define for $\theta =(2n+1)\frac{\pi }{12}$