If `sin alpha=A sin(alpha+beta),Ane0`, then
The value of `tan alpha` is

Options

A. `(Asin beta)/(1-Acosbeta)`
B. `(Asin beta)/(1+A cos beta)`
C. `(A cos beta)/(1-A sin beta)`
D. `(A sin beta)/(1+A cos beta)`

Explanation

To solve the equation \( \sin \alpha = A \sin(\alpha + \beta) \) for \( \tan \alpha \), we will follow these steps:

### Step 1: Apply the sine addition formula
We start with the equation:
\[ \sin \alpha = A \sin(\alpha + \beta) \]
Using the sine addition formula, we can expand \( \sin(\alpha + \beta) \):
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
Substituting this into the equation gives us:
\[ \sin \alpha = A (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \]

### Step 2: Distribute \( A \)
Now, we distribute \( A \):
\[ \sin \alpha = A \sin \alpha \cos \beta + A \cos \alpha \sin \beta \]

### Step 3: Rearrange the equation
Next, we rearrange the equation to isolate terms involving \( \sin \alpha \):
\[ \sin \alpha - A \sin \alpha \cos \beta = A \cos \alpha \sin \beta \]
Factoring out \( \sin \alpha \) from the left side gives:
\[ \sin \alpha (1 - A \cos \beta) = A \cos \alpha \sin \beta \]

### Step 4: Solve for \( \tan \alpha \)
To find \( \tan \alpha \), we divide both sides by \( \cos \alpha \):
\[ \frac{\sin \alpha}{\cos \alpha} (1 - A \cos \beta) = A \sin \beta \]
This simplifies to:
\[ \tan \alpha (1 - A \cos \beta) = A \sin \beta \]
Now, we can isolate \( \tan \alpha \):
\[ \tan \alpha = \frac{A \sin \beta}{1 - A \cos \beta} \]

### Final Answer
Thus, the value of \( \tan \alpha \) is:
\[ \tan \alpha = \frac{A \sin \beta}{1 - A \cos \beta} \]

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