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If $\tan \theta+\tan 4 \theta+\tan 7 \theta=\tan \theta \tan 4 \theta \tan 7 \theta$ then the general solution is
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Verified Answer
The correct answer is:
$\theta=\frac{n \pi}{12}$
We have,
$\tan \theta+\tan 4 \theta+\tan 7 \theta=\tan \theta \tan 4 \theta \tan 7 \theta$
$\tan \theta+\tan 4 \theta=\tan \theta \tan 4 \theta \tan 7 \theta-\tan 7 \theta$
$\tan \theta+\tan 4 \theta=-\tan 7 \theta(1-\tan \theta \tan 4 \theta)$
$\frac{\tan \theta+\tan 4 \theta}{1-\tan \theta \tan 4 \theta}=-\tan 7 \theta$
$\tan (\theta+4 \theta)=-\tan 7 \theta$
$\tan 5 \theta=\tan (-7 \theta)$
$\therefore \quad 5 \theta=-7 \theta$
$12 \theta=0$ or $12 \theta=n \pi$
So, $\quad \theta=\frac{n \pi}{12}, \forall n \in I$
$\tan \theta+\tan 4 \theta+\tan 7 \theta=\tan \theta \tan 4 \theta \tan 7 \theta$
$\tan \theta+\tan 4 \theta=\tan \theta \tan 4 \theta \tan 7 \theta-\tan 7 \theta$
$\tan \theta+\tan 4 \theta=-\tan 7 \theta(1-\tan \theta \tan 4 \theta)$
$\frac{\tan \theta+\tan 4 \theta}{1-\tan \theta \tan 4 \theta}=-\tan 7 \theta$
$\tan (\theta+4 \theta)=-\tan 7 \theta$
$\tan 5 \theta=\tan (-7 \theta)$
$\therefore \quad 5 \theta=-7 \theta$
$12 \theta=0$ or $12 \theta=n \pi$
So, $\quad \theta=\frac{n \pi}{12}, \forall n \in I$
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