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) $ ?

Another approach:
Lets, start by arranging the expression:
$$\tan(20°) \tan(30°) \tan(40°) = \tan(30°) \tan(40°) \tan(20°)$$$$=\tan(30°)  \tan(30°+10°)  \tan(30° - 10°)$$
Now, we will express $\tan(30° + 10°) $ and $\tan(30° - 10°)$ as the ratio of Prosthaphaeresis Formulas, giving us:
  $$\tan(30°) \left( \frac{\tan(30°) + \tan(10°)}{1 - \tan(30°) \tan(10°)}\right)  \left( \frac{\tan(30°) - \tan(10°)}{1 + \tan(30°) \tan(10°)}\right) $$
$$= \tan(30°) \left( \frac{\tan^2(30°) - \tan^2(10°)}{1 - \tan^2(30°) \tan^2(10°)}\right) $$
Substituting the value of $\color{blue}{\tan(30°)}$, 
$$ = \tan(30°)  \left(\frac{1 - 3\tan²(10°)}{3 - \tan²(10°)}\right) $$
Multiplying and dividing by $\color{blue}{\tan(10°)}$,
$$=\tan(30°)  \tan(10°) \left(\frac{1 - 3\tan²(10°)}{3 \tan(10°) - \tan^3(10°)}\right) $$
It can be easily shown that: $\color{blue}{\tan 3A =\large \frac{3 \tan A - \tan^3A}{1-3\tan^2A}}$,
Thus, our problem reduces to: $$=\tan(30°)  \tan(10°) \frac{1}{\tan(3\times 10°)}= \tan(10°)$$
QED!
