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.

We have
$$\tan\left(A-B\right) = \dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} = \dfrac{\tan(B)/2}{1+\dfrac32 \tan^2(B)} = \dfrac{\tan(B)}{2+3\tan^2(B)} = \dfrac{\dfrac{\sin(B)}{\cos(B)}}{2+3 \cdot \dfrac{\sin^2(B)}{\cos^2(B)}}$$
Hence, we have
$$\tan\left(A-B\right) = \dfrac{\sin(B)\cos(B)}{2\cos^2(B)+3\sin^2(B)} = \dfrac{\sin(2B)}{4+2\sin^2(B)} = \dfrac{\sin(2B)}{5-\cos(2B)}$$
where we made use of $\sin(2B) = 2\sin(B) \cos(B)$, $\cos(2B) = 1-2\sin^2(B)$.
