If `3 sin beta=sin(2alpha+beta)`, then `tan (alpha+beta)-2 tan alpha` is

Options

A. independent of `alpha`
B. independent of `beta`
C. dependent of both `alpha` and `beta`.
D. independent of both `alpha` and `beta`.

Explanation

To solve the equation \(3 \sin \beta = \sin(2\alpha + \beta)\) and find the value of \( \tan(\alpha + \beta) - 2 \tan \alpha\), we can follow these steps:

### Step 1: Rewrite the expression for \( \tan(\alpha + \beta) \)

Using the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \]

### Step 2: Substitute the sine and cosine formulas

We can express \( \sin(\alpha + \beta) \) and \( \cos(\alpha + \beta) \) using the sine and cosine addition formulas:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

### Step 3: Substitute into the tangent formula

Now substituting these into the tangent formula:
\[ \tan(\alpha + \beta) = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \]

### Step 4: Write \( \tan(\alpha + \beta) - 2 \tan \alpha \)

Now, we need to express \( \tan \alpha \):
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \]
Thus,
\[ 2 \tan \alpha = \frac{2 \sin \alpha}{\cos \alpha} \]

Now we can write:
\[ \tan(\alpha + \beta) - 2 \tan \alpha = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} - \frac{2 \sin \alpha}{\cos \alpha} \]

### Step 5: Combine the fractions

To combine these fractions, we need a common denominator:
\[ = \frac{(\sin \alpha \cos \beta + \cos \alpha \sin \beta) \cos \alpha - 2 \sin \alpha (\cos \alpha \cos \beta - \sin \alpha \sin \beta)}{\cos \alpha (\cos \alpha \cos \beta - \sin \alpha \sin \beta)} \]

### Step 6: Simplify the numerator

Expanding the numerator:
\[ = \sin \alpha \cos \beta \cos \alpha + \cos \alpha \sin \beta \cos \alpha - 2 \sin \alpha \cos \alpha \cos \beta + 2 \sin^2 \alpha \sin \beta \]

### Step 7: Factor out common terms

Now, we can factor out common terms from the numerator:
\[ = \sin \beta \cos^2 \alpha + \sin \alpha \cos \beta \cos \alpha - 2 \sin \alpha \cos \alpha \cos \beta + 2 \sin^2 \alpha \sin \beta \]

### Step 8: Use the given equation

Since we know that \(3 \sin \beta = \sin(2\alpha + \beta)\), we can use this to simplify further.

### Step 9: Set the expression to zero

Using the given equation, we can set the expression to zero:
\[ 3 \sin \beta - \sin(2\alpha + \beta) = 0 \]

### Conclusion

Thus, we find that:
\[ \tan(\alpha + \beta) - 2 \tan \alpha = 0 \]
This means that the value is independent of both \( \alpha \) and \( \beta \).