Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ 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) $ ?

In a word, yes. You already know that (in degrees) $\tan 20\cdot\tan30=\tan10\cdot\tan50$ so
$$\tan20\cdot\tan30\cdot\tan40 = \tan10\cdot\tan50\cdot\tan40$$
and your observation that 
$$\frac{1}{\tan40}=\tan50$$
is all you need.
