Question

A solution of $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$ is given by $\left(n\in Z\right)$

• A. $n\pi$
• B. $\left(2n+1\right)\frac{\pi }{4}$
• C. $n\pi +{\left(-1\right)}^n\frac{\pi }{12}$
• D. $\frac{n\pi }{12}$

Answer 1

The correct option is D $\frac{n\pi }{12}$

Let $A+B+C=n\pi ,n\in Z$

$\tan (A+B+C)=\tan n\pi =0$

$⟹\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1+\tan A\tan B+\tan B\tan C+\tan C\tan A}=0$

$⟹\tan A+\tan B+\tan C=\tan A\tan B\tan C$

So given $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$

$⟹\theta +4\theta +7\theta =n\pi$

$⟹12\theta =n\pi$

$⟹\theta =\frac{n\pi }{12}$