Trigonometric Ratios Of Multiple Angles If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$
I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.

Notice, here is another method: $$RHS=\frac{\sin 2B}{5-\cos 2B}$$
$$=\frac{\frac{2\tan B}{1+\tan^2B}}{5-\frac{1-\tan^2 B}{1+\tan^2A}}$$
$$=\frac{2\tan B}{4+6\tan^2B}$$
$$=\frac{\tan B}{2+3\tan^2B}$$
$$=\frac{(3\tan B)-2\tan B}{2+(3\tan B)\tan B}$$
setting $3\tan B=2\tan A$, 
$$=\frac{2\tan A-2\tan B}{2+2\tan A\tan B}$$
$$=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
$$=\tan(A-B)=LHS$$
