Given :
$2\tan\alpha=3\tan\beta$
$\tan\alpha=\large\frac{3}{2}$$\tan \beta$
$\tan(\alpha-\beta)=\large\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$\Rightarrow \large\frac{\Large\frac{3}{2}\tan\beta-\tan\beta}{1+\Large\frac{3}{2}\tan^2\beta}$
$\Rightarrow \large\frac{\tan\beta}{2+3\tan^2\beta}$
$\Rightarrow \large\frac{\Large\frac{\sin\beta}{\cos\beta}}{2+3\Large\frac{\sin^2\beta}{\cos^2\beta}}$
$\Rightarrow \large\frac{\sin\beta\cos\beta}{2\cos^2\beta+3\sin^2\beta}$
$\Rightarrow \large\frac{\sin \beta\cos\beta}{1+\cos 2\beta+\Large\frac{3}{2}\normalsize(1+\cos 2\beta)}$
$\Rightarrow \large\frac{2\sin \beta\cos\beta}{5+2\cos \beta-3\cos 2\beta}$
$\Rightarrow \large\frac{\sin2 \beta}{5-\cos 2\beta}$=LHS
Hence proved.