Question

If $\tan \theta +\tan 4\theta +\tan 7\theta =\tan \theta \tan 4\theta \tan 7\theta$, then $\theta =$

• A. $\theta =\frac{n\pi }{4}$
• B. $\theta =\frac{n\pi }{7}$
• C. $\theta =\frac{n\pi }{12};n\ne 6(2k+1)(n,k\in I)$
• D. $\theta =n\pi$

Answer 1

The correct option is C $\theta =\frac{n\pi }{12};n\ne 6(2k+1)(n,k\in I)$

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

let $A=\theta ,B=4\theta ,C=7\theta$

then numerator $\tan \theta +\tan 4\theta +\tan 7\theta -\tan \theta \tan 4\theta \tan 7\theta$

So to solve if numberator $=0$

then $\tan (\theta +4\theta +7\theta )=\tan 12\theta =0$

or $12\theta =n\pi$

$\theta =\frac{11\pi }{12}$