$\sin B=\large\frac{1}{5}$$\sin(2A+B)$
$\Rightarrow \large\frac{\sin(2A+B)}{\sin B}$$=5$
$\Rightarrow \large\frac{\sin(2A+B)+\sin B}{\sin(2A+B)-\sin B}$$=\large\frac{6}{4}$
$\Rightarrow \large\frac{\sin(A+B)\cos A}{\cos(A+B)\sin A}=\frac{3}{2}$
$\Rightarrow \large\frac{\tan(A+B)}{\tan A}=\frac{3}{2}$
Hence (c) is the right option.