If cos (A−B)=

$\cos (A-B)=\frac{3}{5}$ Given

and $\tan A \tan B=2$

Since $\cos (A-B)=3 / 5$

i. e. $\cos A \cos B+\sin A \sin B=3 / 5$

i. e. $\cos A \cos B\left(1+\frac{\sin A \sin B}{\cos A \cos B}\right)=3 / 5$

i. e. $\cos A \cos B(1+\tan A \tan B)=3 / 5$

i.e. $\cos A \cos B(1+2)=3 / 5$

$\cos A \cos B=\frac{3}{5} \times \frac{1}{3}$

$\cos A \cos B=\frac{1}{5}$

$\therefore \sin A \sin B=\frac{3}{5}-\frac{1}{5}$

$\sin A \sin B=\frac{2}{5}$