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) $ ?
 A: 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.
A: This may be the hard way, but you can use $$\tan\theta={e^{i\theta}-e^{- i\theta}\over i(e^{i\theta}+e^{-i\theta})}$$ to write everything in terms of $\zeta=e^{\pi i/18}$. Then $\zeta$ is a primitive 36th root of unity, and you can use that to make simplifications. 
EDIT: In light of the comments, I flesh this out a bit. 
We start with Euler's formula, $$e^{ix}=\cos x+i\sin x\tag1$$ We'll say a bit more about where this comes from, later. Replace $x$ everywhere with $-x$, and use $\cos(-x)=\cos x$ and $\sin(-x)=-\sin x$, to get $$e^{-ix}=\cos x-i\sin x\tag2$$ Add (1) and (2), and divide by 2 to get $$\cos x={e^{ix}+e^{-ix}\over2}\tag3$$ Subtract (2) from (1) and divide by $2i$ to get $$\sin x={e^{ix}-e^{-ix}\over2i}\tag4$$ Divide (4) by (3) to get $$\tan x={e^{ix}-e^{-ix}\over i(e^{ix}+e^{-ix})}\tag5$$ 10 degrees is $\pi/18$ radians; put in $x=\pi/18$ to get $$\tan{\pi\over18}={e^{i{\pi\over18}}-e^{-i{\pi\over18}}\over i(e^{i{\pi\over18}}+e^{-i{\pi\over18}})}\tag6$$ Let $\zeta=e^{\pi i/18}$; then $$\tan{\pi\over18}={\zeta-\zeta^{-1}\over i(\zeta+\zeta^{-1})}\tag7$$ Similarly, 20, 30, 40 degrees are $2\pi/18,3\pi/18,4\pi/18$ radians, respectively, and we get $$\tan{2\pi\over18}={\zeta^2-\zeta^{-2}\over i(\zeta^2+\zeta^{-2})},\quad\tan{3\pi\over18}={\zeta^3-\zeta^{-3}\over i(\zeta^3+\zeta^{-3})},\quad\tan{4\pi\over18}={\zeta^4-\zeta^{-4}\over i(\zeta^4+\zeta^{-4})}\tag8$$ So now you can take the equation in the title, and write it completely in terms of these powers of $\zeta$, and when you multiply through by all the denominators and an appropriate power of $\zeta$, you just get an equation in powers of $\zeta$ that needs to be verified. 
At that point, you may need some properties of $\zeta$. By (1), $\zeta^{36}=1$, and $\zeta^{18}=-1$. You may need some more properties of $\zeta$, but we'd have to see the equation first. 
And, as suggested in the first line, and verified by Rick's solution, this is the hard way, for this problem. But it's still good to know these methods, as there are situations where they provide the easiest way. 
A: 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!
A: Due to $\tan 40^\circ*\tan 20^\circ$
$$=\frac{\sin 40^\circ*\sin 20^\circ}{\cos 40^\circ*\cos 20^\circ}$$
$$=\frac{4\sin 40^\circ*(\sin 20^\circ)^2}{4\cos 40^\circ*\cos 20^\circ*\sin 20^\circ}$$
$$=\frac{2\sin 40^\circ*(1-\cos 40^\circ)}{2\cos 40^\circ*\sin 40^\circ}$$
$$=\frac{2\sin 40^\circ-2\sin 40^\circ*\cos 40^\circ}{\sin 80^\circ}$$
$$=\frac{2\sin 40^\circ-\sin 80^\circ}{\cos10^\circ}$$
$$=\frac{\sin 40^\circ-(\sin 80^\circ-\sin 40^\circ)}{\cos10}$$
$$=\frac{\sin 40^\circ-2\cos 60^\circ*\sin 20^\circ}{\cos10}$$
$$=\frac{\sin 40^\circ-\sin 20^\circ}{\cos10}$$
$$=\frac{2\cos 30^\circ*\sin 10^\circ}{\cos10}$$
$$=\cot 30^\circ*\tan 10^\circ$$
Hence, 
$$\tan 20^\circ*\tan 30^\circ*\tan 40^\circ$$
$$=\tan 30^\circ*\cot 30^\circ*\tan 10^\circ$$
$$=\tan 10^\circ$$
A: Using Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
$$\tan20^\circ\tan(60^\circ-20^\circ)\tan(60^\circ+20^\circ)=\tan(3\cdot20^\circ)$$
and $\tan80^\circ=\cot(90^\circ-80^\circ)=\dfrac1{\tan?}$
