Question: The question is: 

> 
> Prove $$\tan70^\circ - \tan50^\circ + \tan10^\circ = \sqrt{3}.$$
> 
> 
> 

I had no idea how to do it & proceeded by making RHS = $\tan60^\circ$ but this doesn't make any help. Please help me solving this.

Answer: 
$\tan3x=\dfrac{3t-t^3}{1-3t^2}$ where $t=\tan x$

If $\tan3x=\tan3A$

$\implies t^3-3t^2\tan3A-3t+\tan3A=0$

$\implies\sum\_{r=-1}^1\tan(60^\circ r+x)=\dfrac{3\tan3A}1$

Observe that $\tan3x=\tan30^\circ$ for $x=70^\circ,-50^\circ,10^\circ$

$\implies3x=180^\circ n+30^\circ$ where $n$ is any integer

$x=60^\circ n+10^\circ;n=-1,0,1$
You're right, @André Nicolas:
$$\tan70° - \tan50° + \tan10° = \tan(60°+10°)-\tan(60°-10°) +\tan10°=$$ $$ =\frac{\sqrt{3}+\tan10°}{1-\sqrt{3}\cdot\tan10°}+\frac{\sqrt{3}-\tan10°}{1+\sqrt{3}\cdot\tan10°}+\tan10°= 3\frac{\tan10°\cdot(3-\tan^210°)}{1-3\cdot\tan^210°}=$$ $$=3\cdot\tan3\cdot10° =3\cdot\frac{1}{\sqrt{3}}=\sqrt{3}.$$