Calculate without using tables:
$\tan 20^\circ \tan 40^\circ \tan 80^\circ$
The given expression is: $\;\;$ $\tan 20^\circ \tan 40^\circ \tan 80^\circ$
$= \dfrac{\sin 20^\circ \sin 40^\circ \sin 80^\circ}{\cos 20^\circ \cos 40^\circ \cos 80^\circ}$
$= \dfrac{\sin 20^\circ \times \dfrac{1}{2} \left[2 \sin 80^\circ \sin 40^\circ\right]}{\cos 20^\circ \times \dfrac{1}{2} \left[2 \cos 80^\circ \cos 40^\circ\right]}$
$= \dfrac{\sin 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) - \cos \left(80^\circ + 40^\circ\right)\right]}{\cos 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) + \cos \left(80^\circ + 40^\circ\right)\right]}$
$= \dfrac{\sin 20^\circ \left[\cos 40^\circ - \cos 120^\circ\right]}{\cos 20^\circ \left[\cos 40^\circ + \cos 120^\circ\right]}$
$= \dfrac{\sin 20^\circ \left[\cos 40^\circ - \left(\dfrac{-1}{2}\right)\right]}{\cos 20^\circ \left[\cos 40^\circ + \left(\dfrac{-1}{2}\right)\right]}$
$= \dfrac{2 \sin 20^\circ \cos 40^\circ + \sin 20^\circ}{2 \cos 20^\circ \cos 40^\circ - \cos 20^\circ}$
$= \dfrac{\sin \left(40^\circ + 20^\circ\right) + \sin \left(20^\circ - 40^\circ\right) + \sin 20^\circ}{\cos \left(40^\circ - 20^\circ\right) + \cos \left(40^\circ + 20^\circ\right) - \cos 20^\circ}$
$= \dfrac{\sin 60^\circ + \sin \left(-20^\circ\right) + \sin 20^\circ}{\sin 20^\circ + \cos 60^\circ - \cos 20^\circ}$
$= \dfrac{\sin 60^\circ - \sin 20^\circ + \sin 20^\circ}{\cos 60^\circ}$
$= \dfrac{\sin 60^\circ}{\cos 60^\circ}$
$= \tan 60^\circ = \sqrt{3}$