Question

If $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$, then the general solution is

• A. $\theta =\frac{n\pi }{4}$
• B. $\theta =\frac{n\pi }{12}$
• C. $\theta =\frac{n\pi }{6}$
• D. None of these

Answer 1

Answer: (B)

$\theta =\frac{n\pi }{12}$

Given : $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$ .. $\left(i\right)$

We know that, $\tan (a+b+c)=\frac{\tan a+\tan b+\tan c-\tan a\tan b\tan c}{1-\tan a\tan b-\tan b\tan c-\tan a\tan c}$ .... $\left(ii\right)$

Now, $\tan \left(12\theta \right)=\tan (\theta +4\theta +7\theta )$

Then by using $\left(ii\right)$, we have

$\tan \left(12\theta \right)=\frac{\tan \theta +\tan 4\theta +\tan 7\theta -\tan \theta \tan 4\theta \tan 7\theta }{1-\tan \theta \tan 4\theta -\tan \theta \tan 7\theta -\tan 4\theta \tan 7\theta }=0$

$⟹\tan \left(12\theta \right)=0$ ... From $\left(i\right)$

$⟹12\theta =n\pi$

$⟹\theta =\frac{n\pi }{12}$

Hence, B is correct.