Question:
If $\frac{\sin (2A + B)}{\sin B} = 5,$ then find $\frac{\tan (A + B)}{\tan A}.$

Answer:
From $\frac{\sin (2A + B)}{\sin B} = 5,$
\[\sin (2A + B) = 5 \sin B.\]We can write this as $\sin (A + (A + B)) = 5 \sin ((A + B) - A),$ so from the angle addition and subtraction formula,
\[\sin A \cos (A + B) + \cos A \sin (A + B) = 5 \sin (A + B) \cos A - 5 \cos (A + B) \sin A.\]Then
\[6 \sin A \cos (A + B) = 4 \sin (A + B) \cos A,\]so
\[\frac{\sin (A + B) \cos A}{\cos (A + B) \sin A} = \frac{3}{2}.\]In other words,
\[\frac{\tan (A + B)}{\tan A} = \boxed{\frac{3}{2}}.\]