If 2 tan α=3 tan β,

Given:

$2 \tan \alpha=3 \tan \beta$

$\mathrm{LHS}=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \times \tan \beta}$

$=\frac{\frac{\frac{3}{2} \times \tan \beta-\tan \beta}{1+\frac{3}{2} \tan ^{2} \beta}}{(\because 2 \tan \alpha=3 \tan \beta)}$

$=\frac{\frac{1}{2} \times \tan \beta}{1+\frac{3}{2} \tan ^{2} \beta}=\frac{\tan \beta}{2+3 \tan ^{2} \beta}$

$=\frac{\frac{1}{2} \times \tan \beta}{1+\frac{3}{2} \tan ^{2} \beta}=\frac{\tan \beta}{2+3 \tan ^{2} \beta}$

$=\frac{\frac{\sin \beta}{\cos \beta}}{2+3 \frac{\sin ^{2} \beta}{\cos ^{2} \beta}}=\frac{\frac{\sin \beta}{\cos \beta} \times \cos ^{2} \beta}{2 \cos ^{2} \beta+3 \sin ^{2} \beta}$

$=\frac{\sin \beta \times \cos \beta}{2 \cos ^{2} \beta+2 \sin ^{2} \beta+\sin ^{2} \beta}$

$=\frac{1}{2} \frac{2 \sin \beta \times \cos \beta}{2\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\sin ^{2} \beta}$

$=\frac{1}{2} \frac{\sin 2 \beta}{\left(2+\sin ^{2} \beta\right)}=\frac{\sin 2 \beta}{4+2 \sin ^{2} \beta}$

$=\frac{\sin 2 \beta}{4+2\left(1-\cos ^{2} \beta\right)}=\frac{\sin 2 \beta}{6-2 \cos ^{2} \beta}$

$=\frac{\sin 2 \beta}{6-(1+\cos 2 \beta)} \quad\left(\because 1+\cos 2 \beta=2 \cos ^{2} \beta\right)$

$=\frac{\sin 2 \beta}{5-\cos 2 \beta}=\mathrm{RHS}$

Hence proved.