how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?

I know how to prove
 $ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $

in this way:

$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$

$\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$

$\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$

$\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$

$\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$

$\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$ 

Could this help to prove the first one and how ?Do i just need to know that $ \frac{1}{\tan\theta}=\tan(90^{\circ}-\theta) $ ?