Question

Prove that $\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +\frac{2\pi }{3})=3$.

Answer 1

$\tan \frac{\pi }{3}=\sqrt{3},\tan \frac{2\pi }{3}=\tan (\pi -\frac{\pi }{3})$
$=-\tan \frac{\pi }{3}=-\sqrt{3}$
Now $\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +2\frac{2\pi }{3})$
$=\tan \theta +\frac{\tan \theta +\sqrt{3}}{1-\sqrt{\left(3\right)}\tan \theta }+\frac{\tan \theta -\sqrt{3}}{1+\sqrt{\left(3\right)}\tan \theta }$
$=\frac{\tan \theta (1-3{\tan }^2\theta )+(\tan \theta +\sqrt{3})(1+\sqrt{3}\tan \theta )+(\tan \theta -\sqrt{3})(1-\sqrt{3}\tan \theta )}{1-3{\tan }^2\theta }$
$=\frac{\tan \theta -3{\tan }^2\theta +2\tan \theta +\sqrt{3}\left(2\sqrt{3}\tan \theta \right)}{1-3{\tan }^2\theta }$
$=\frac{9\tan \theta -3{\tan }^3\theta }{1-3{\tan }^2\theta }$
$=3\cdot \frac{(3\tan \theta -{\tan }^3\theta )}{1-3{\tan }^2\theta }=3\tan 3\theta$.
Hence the given equation reduces to $3\tan 3\theta =3$
$\therefore \tan 3\theta =1=\tan \frac{\pi }{4}$
$\therefore 3\theta =n\pi +\frac{\pi }{4}$
or $\theta =\frac{n\pi }{3}+\frac{\pi }{12}$