Question

Solve: $\tan \theta +\tan 2\theta +\tan 3\theta =0$.

Answer 1

$\tan \theta +\tan 2\theta +\tan (\theta +2\theta )=0$

$(\tan \theta +\tan 2\theta )+\frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta }=0$

$(\tan \theta +\tan 2\theta )(1-\tan \theta \tan 2\theta +1)=0$

$\tan 2\theta +\tan \theta =0$

$\therefore \tan 2\theta =-\tan \theta =\tan (-\theta )$

$\therefore 2\theta =n\pi -\theta$ or $3\theta =n\pi$

$\therefore \theta =n\pi /3$

From $2$nd factor

$\tan \theta \tan 2\theta =2$ or $\tan \theta \cdot \frac{2\tan \theta }{1-{\tan }^2\theta }=2$

${\tan }^2\theta =1-{\tan }^2\theta$ or $2{\tan }^2\theta =1$

$\therefore \tan \theta =\pm \frac{1}{\sqrt{\left(2\right)}}=\pm \tan \alpha$ where $\tan \alpha =\frac{1}{\sqrt{\left(2\right)}}$

$\therefore \theta =n\pi \pm \alpha$ where $\alpha ={\tan }^{-1}\frac{1}{\sqrt{\left(2\right)}}$

$0<\alpha <\pi /2$