<s>[INST] 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... 
 [/INST] Variation of lab bhattacharjee's answer:
$$\tan^2x=\tan(x-a)·\tan(x-b) \Rightarrow \\
\tan^2x+1=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\frac{1}{\cos^2 x}=\tan(x-a)·\tan(x-b)+1 \Rightarrow \\
\cos^2x =\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\
\frac{1+\cos 2x}{2}=\frac1{\tan(x-a)·\tan(x-b)+1} \Rightarrow \\\\
\cos 2x=\frac2{\tan(x-a)·\tan(x-b)+1}-1 \Rightarrow \\
\cos 2x=\frac{1-\tan (x-a)\cdot \tan (x-b)}{1+\tan (x-a)\cdot \tan (x-b)}=\\
\cos 2x=\frac{\cos (x-a)\cos (x-b)-\sin (x-a)\cos (x-b)}{\cos (x-a)\cos (x-b)+\sin (x-a)\cos (x-b)} \Rightarrow \\
\cos 2x=\frac{\cos (2x-(a+b))}{\cos (a-b)} \Rightarrow \\
\cos 2x=\frac{\cos 2x\cos (a+b)+\sin 2x\sin (a+b)}{\cos (a-b)}\Rightarrow \\
\cos 2x\cos (a-b)=\cos 2x\cos (a+b)+\sin 2x\sin (a+b) \Rightarrow \\
\cos 2x[\cos (a-b)-\cos (a+b)]=\sin 2x\sin (a+b)\Rightarrow \\
\tan 2x=\frac{\cos (a-b)-\cos(a+b)}{\sin (a+b)}=\\
\frac{2\sin a\sin b}{\sin(a+b)}.$$
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