A formula for tan(2x) Help with solving...

Suppose that $\tan^2x=\tan(x-a)·\tan(x-b)$, show that $$\tan(2x)=\frac{2\sin(a)·\sin(b)}{\sin(a+b)}$$

As far as I know, so far the $\tan2x$ can be converted to $\frac{2\tan x}{1-\tan^2x}$ using the double angle formula and the $\tan 2x$ can be further be substituted to the following given item ($\tan(x-a·)\tan(x-b)$) however the problem now is that there are no ways to simplify this to my knowledge... 

Hint: Using the addition formulas for the tan-function and factorizing $$\tan(x-a)\tan(x-b)-\tan^2(x)=0$$ we get
$$- \left(  \left( \tan \left( x \right)  \right) ^{2}+1 \right) 
 \left(  \left( \tan \left( x \right)  \right) ^{2}\tan \left( a
 \right) \tan \left( b \right) +\tan \left( x \right) \tan \left( a
 \right) +\tan \left( x \right) \tan \left( b \right) -\tan \left( a
 \right) \tan \left( b \right)  \right)=0$$ Can you solve this equation?
