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... 

The intial equations can be written
$$t^2=\frac{t-t_a}{1+t\,t_a}\frac{t-t_b}{1+t\,t_b},$$
$$\frac{2t}{1-t^2}=\frac{2t_at_b}{t_a+t_b}.$$
From the second equation, we can draw $t^2$ as a linear function of $t$. Then, replacing $t^2$ several times in the first equation, we should get an identity.
With $c$ denoting the cotangent,
$$t^2=1-(c_a+c_b)t,$$ and 
$$t^2=\frac{t-t_a}{1+t\,t_a}\frac{t-t_b}{1+t\,t_b}$$
becomes
$$1-(c_a+c_b)t=\frac{1-(c_a+c_b)t-(t_a+t_b)t+t_at_b}{1+(t_a+t_b)t+t_at_b(1-(c_a+c_b)t)}.$$
The denominator simplifies as 
$$1+t_at_b,$$
and we do have
$$(1+t_at_b)(c_a+c_b)=c_a+c_b+t_a+t_b.$$
