Question

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

Answer 1

From \(\frac{\sin (2 A+B)}{\sin B}=5\)
\[
\sin (2 A+B)=5 \sin B
\]
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,
\[
\frac{\tan (A+B)}{\tan A}=\frac{3}{2}
\]

Final Answer: The final answer is \(\frac{3}{2}\)