If `sinB=1/5sin(2A+B), then (tan(A+B))/(tanA)` is equal to

Options

A. `5//3`
B. `2//3`
C. `3//2`
D. `3//5`

Answer

To solve the problem, we start with the given equation:

\[ \sin B = \frac{1}{5} \sin(2A + B) \]

### Step 1: Rewrite the equation
Multiply both sides by 5 to eliminate the fraction:

\[ 5 \sin B = \sin(2A + B) \]

### Step 2: Use the sine addition formula
We can use the sine addition formula to expand \(\sin(2A + B)\):

\[ \sin(2A + B) = \sin(2A) \cos(B) + \cos(2A) \sin(B) \]

Substituting this back into our equation gives us:

\[ 5 \sin B = \sin(2A) \cos(B) + \cos(2A) \sin(B) \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:

\[ 5 \sin B - \cos(2A) \sin(B) = \sin(2A) \cos(B) \]

Factoring out \(\sin B\) from the left side:

\[ \sin B (5 - \cos(2A)) = \sin(2A) \cos(B) \]

### Step 4: Divide both sides by \(\sin A\)
Now, we want to find \(\frac{\tan(A + B)}{\tan A}\). We can express \(\tan(A + B)\) using the tangent addition formula:

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Thus, we have:

\[ \frac{\tan(A + B)}{\tan A} = \frac{\tan A + \tan B}{\tan A(1 - \tan A \tan B)} \]

### Step 5: Express \(\tan B\) in terms of \(\tan A\)
From the equation \(5 \sin B = \sin(2A + B)\), we can relate \(\tan B\) to \(\tan A\). Using the identity \(\sin(2A) = 2 \sin A \cos A\), we can express \(\sin(2A)\) and \(\cos(2A)\) in terms of \(\tan A\):

\[ \sin(2A) = \frac{2 \tan A}{1 + \tan^2 A}, \quad \cos(2A) = \frac{1 - \tan^2 A}{1 + \tan^2 A} \]

### Step 6: Substitute and simplify
Substituting these into our equation gives us a relationship between \(\tan A\) and \(\tan B\). After some algebra, we find:

\[ \frac{\tan(A + B)}{\tan A} = \frac{3}{2} \]

### Conclusion
Thus, the final answer is:

\[ \frac{\tan(A + B)}{\tan A} = \frac{3}{2} \]