Let `alpha, betain(0,pi/2),3sin(alpha+beta)=2sin(alpha-beta)` and `tanalpha=ktanbeta` then the value of k is

Options

A. `-5`
B. 4
C. 3
D. `-3`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ 3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta) \]
and
\[ \tan \alpha = k \tan \beta \]

2. **Rearranging the First Equation**:
We can rearrange the first equation:
\[ \frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = \frac{2}{3} \]

3. **Using the Sine Addition and Subtraction Formulas**:
Using the sine addition and subtraction formulas:
\[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]
\[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]

We substitute these into our equation:
\[ \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \frac{2}{3} \]

4. **Cross Multiplying**:
Cross multiplying gives:
\[ 3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta) \]

5. **Expanding and Rearranging**:
Expanding both sides:
\[ 3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta \]
Rearranging gives:
\[ 3 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta = -2 \cos \alpha \sin \beta - 3 \cos \alpha \sin \beta \]
Simplifying:
\[ \sin \alpha \cos \beta = -5 \cos \alpha \sin \beta \]

6. **Dividing by \(\cos \alpha \sin \beta\)** (assuming they are non-zero):
\[ \frac{\sin \alpha}{\sin \beta} = -5 \frac{\cos \alpha}{\cos \beta} \]

7. **Using the Tangent Identity**:
From the tangent identity, we have:
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha}, \quad \tan \beta = \frac{\sin \beta}{\cos \beta} \]
Thus:
\[ \frac{\tan \alpha}{\tan \beta} = -5 \implies \tan \alpha = -5 \tan \beta \]

8. **Substituting into the Second Equation**:
Since \(\tan \alpha = k \tan \beta\), we can equate:
\[ -5 \tan \beta = k \tan \beta \]
Dividing both sides by \(\tan \beta\) (assuming \(\tan \beta \neq 0\)):
\[ k = -5 \]

9. **Final Answer**:
Therefore, the value of \(k\) is:
\[ \boxed{-5} \]