Question

$tan{20}^0+tan{40}^0+\sqrt{3}tan{20}^{0}tan{40}^0$ is equal to:-

Answer 1

We have $20+40=60$

So tan $60=\tan (20+40)$

$=\frac{\left(\tan \right(20)+\tan (40)}{1-\tan 20\tan 40}$

or

$\tan 60-\tan 60\tan 20\tan 40=\tan 20+\tan 40$

or

$\sqrt{3}-\sqrt{3}\tan 20\tan 40=\tan 20+\tan 40$

or

$\tan 20+\tan 40+\sqrt{3}\tan 20\tan 40=\sqrt{3}$.