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

Options

A. `(sinalpha(1+A cos beta))/(A cos alpha cos beta)`
B. `(sinalpha(1-A cos beta))/(A cos alpha cos beta)`
C. `(cos (1-A sinbeta))/(A cos alpha cos beta)`
D. `(cos alpha(1+A sin beta))/(A cos alpha cos beta)`

Explanation

To solve the equation \( \sin \alpha = A \sin(\alpha + \beta) \) where \( A \neq 0 \), we will follow these steps:

### Step 1: Apply the sine addition formula
We start with the given 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 our equation gives:
\[ \sin \alpha = A (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \]

### 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, we have:
\[ \sin \alpha (1 - A \cos \beta) = A \cos \alpha \sin \beta \]

### Step 4: Solve for \( \tan \alpha \)
Now, we can express \( \tan \alpha \):
\[ \tan \alpha = \frac{A \sin \beta}{1 - A \cos \beta} \]

### Step 5: Express \( \tan \beta \)
To find \( \tan \beta \), we can use the relationship between \( \sin \beta \) and \( \cos \beta \):
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} \]
From our previous equation, we can express \( \sin \beta \) in terms of \( \tan \alpha \):
\[ \sin \beta = \frac{\tan \alpha (1 - A \cos \beta)}{A} \]
Substituting this into the equation for \( \tan \beta \):
\[ \tan \beta = \frac{\frac{\tan \alpha (1 - A \cos \beta)}{A}}{\cos \beta} \]
This simplifies to:
\[ \tan \beta = \frac{\tan \alpha (1 - A \cos \beta)}{A \cos \beta} \]

### Conclusion
At this point, we have expressed \( \tan \beta \) in terms of \( \tan \alpha \) and \( \cos \beta \). However, we can also derive a more direct relationship based on the rearranged equation.

The final expression for \( \tan \beta \) can be simplified further depending on the context or specific values of \( A \), \( \alpha \), and \( \beta \).