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... 
 A: We have $$\dfrac{\tan^2x}1=\dfrac{\sin(x-a)\sin(x-b)}{\cos(x-a)\cos(x-b)}$$
Apply https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm,
$$\dfrac{1-\tan^2x}{1+\tan^2x}=\dfrac{\cos(2x-(a+b))}{\cos(b-a)}$$
$$\implies\cos(b-a)\cos2x=\cos2x\cos(a+b)+\sin2x\sin(a+b)$$
$$\sin2x(\cdots)=\cos2x(\cdots)$$
A: 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
$$(1+t_at_b)(c_a+c_b)=c_a+c_b+t_a+t_b.$$
A: 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?
A: Variation of lab bhattacharjee's answer:
$$\tan^2x=\tan(x-a)·\tan(x-b) \Rightarrow \\
\tan^2x+1=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\frac{1}{\cos^2 x}=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\cos^2x =\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\
\frac{1+\cos 2x}{2}=\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\\\
\cos 2x=\frac2{\tan(x-a)·\tan(x-b)+1}-1 \Rightarrow \\
\cos 2x=\frac{1-\tan (x-a)\cdot \tan (x-b)}{1+\tan (x-a)\cdot \tan (x-b)}=\\
\cos 2x=\frac{\cos (x-a)\cos (x-b)-\sin (x-a)\cos (x-b)}{\cos (x-a)\cos (x-b)+\sin (x-a)\cos (x-b)} \Rightarrow \\
\cos 2x=\frac{\cos (2x-(a+b))}{\cos (a-b)} \Rightarrow \\
\cos 2x=\frac{\cos 2x\cos (a+b)+\sin 2x\sin (a+b)}{\cos (a-b)}\Rightarrow \\
\cos 2x\cos (a-b)=\cos 2x\cos (a+b)+\sin 2x\sin (a+b) \Rightarrow \\
\cos 2x[\cos (a-b)-\cos (a+b)]=\sin 2x\sin (a+b)\Rightarrow \\
\tan 2x=\frac{\cos (a-b)-\cos(a+b)}{\sin (a+b)}=\\
\frac{2\sin a\sin b}{\sin(a+b)}.$$
