If cos(A - B) = 3 / 5 and tanA tanB = 2, then

$\cos (\mathrm{A}-\mathrm{B})=\frac{3}{5}\text{ or }5\cos \mathrm{Acos}\mathrm{B}+5\sin \mathrm{Asin}\mathrm{B}=3$
From the second realtion,
we have sinAsinB=2cosAcosB ....(ii)
$$\begin{gathered} 5\cos \mathrm{Acos}\mathrm{B}+5\sin \mathrm{Asin}\mathrm{B}=3 \\ \Rightarrow 5\cos \mathrm{Acos}\mathrm{B}+5(2\cos \mathrm{Acos}\mathrm{B})=3 \\ \Rightarrow 15\cos \mathrm{Acos}\mathrm{B}=3 \\ \Rightarrow \cos \mathrm{Acos}\mathrm{B}=\frac{1}{5}\text{ and }\sin \mathrm{Asin}\mathrm{B}=\frac{2}{5} \end{gathered}$$