If $\sin B=\large\frac{1}{5}$$\sin(2A+B)$ then $\large\frac{\tan(A+B)}{\tan A}$ is equal to

Question

$(a)\;\large\frac{5}{3}$$\qquad(b)\;\large\frac{2}{3}$$\qquad(c)\;\large\frac{3}{2}$$\qquad(d)\;\large\frac{3}{5}$

Answer 1

$\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.