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If $\frac{\sin (2 A+B)}{\sin B}=5$, then find $\frac{\tan (A+B)}{\tan A}$.
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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 $\frac{3}{2}$
by Diamond (71,587 points)

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