If `bsinalpha=asin(alpha+2beta)`, then `(a+b)/(a-b)=?`

Options

A. `(tanbeta)/(tan(alpha+beta))`
B. `(cotbeta)/(cot(alpha-beta))`
C. `(cotbeta)/(cot(alpha+beta))`
D. `(tanbeta)/(tan(alpha-beta))`

Answer

To solve the equation \( b \sin \alpha = a \sin(\alpha + 2\beta) \) and find the value of \( \frac{a+b}{a-b} \), we can follow these steps:

### Step 1: Rewrite the Given Equation
We start with the equation:
\[ b \sin \alpha = a \sin(\alpha + 2\beta) \]

### Step 2: Use the Sine Addition Formula
We apply the sine addition formula:
\[ \sin(\alpha + 2\beta) = \sin \alpha \cos 2\beta + \cos \alpha \sin 2\beta \]
Substituting this into the equation gives:
\[ b \sin \alpha = a (\sin \alpha \cos 2\beta + \cos \alpha \sin 2\beta) \]

### Step 3: Rearranging the Equation
Rearranging the equation, we can factor out \( \sin \alpha \):
\[ b \sin \alpha = a \sin \alpha \cos 2\beta + a \cos \alpha \sin 2\beta \]
This can be rewritten as:
\[ b \sin \alpha - a \sin \alpha \cos 2\beta = a \cos \alpha \sin 2\beta \]
Factoring out \( \sin \alpha \) from the left side gives:
\[ \sin \alpha (b - a \cos 2\beta) = a \cos \alpha \sin 2\beta \]

### Step 4: Divide Both Sides
Assuming \( \sin \alpha \neq 0 \), we can divide both sides by \( \sin \alpha \):
\[ b - a \cos 2\beta = a \frac{\cos \alpha \sin 2\beta}{\sin \alpha} \]
This leads us to:
\[ \frac{b - a \cos 2\beta}{a \frac{\cos \alpha \sin 2\beta}{\sin \alpha}} = 1 \]

### Step 5: Componendo and Dividendo
Now, we apply the concept of componendo and dividendo:
\[ \frac{\sin \alpha + \sin(\alpha + 2\beta)}{\sin \alpha - \sin(\alpha + 2\beta)} = \frac{a + b}{a - b} \]
Using the sine addition formula again, we can simplify this expression.

### Step 6: Simplifying the Expression
Using the sine addition and subtraction formulas:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
\[ \sin x - \sin y = 2 \cos\left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2}\right) \]
Applying these formulas, we can simplify the expression on the left side.

### Step 7: Final Expression
After simplification, we find:
\[ \frac{a+b}{a-b} = \frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} \]

### Conclusion
Thus, the final answer is:
\[ \frac{a+b}{a-b} = \frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} \]