If `"sin" alpha=ksinbeta` then `tan((alpha-beta)/(2)).cot((alpha+beta)/(2))` is equal to

Options

A. (a)`(k-1)/(k+1)`
B. (b)`(k+1)/(k-1)`
C. (c)`(1+k)/(1-k)`
D. (d)`(1-k)/(1+k)`

Answer

To solve the problem, we need to find the value of \( \tan\left(\frac{\alpha - \beta}{2}\right) \cot\left(\frac{\alpha + \beta}{2}\right) \) given that \( \sin \alpha = k \sin \beta \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \sin \alpha = k \sin \beta \]
This implies:
\[ \frac{\sin \alpha}{\sin \beta} = k \]

2. **Apply the component to a dividendo:**
Using the component to a dividendo method, we have:
\[ \frac{\sin \alpha - \sin \beta}{\sin \alpha + \sin \beta} = \frac{k - 1}{k + 1} \]

3. **Use the sine subtraction and addition formulas:**
We know:
\[ \sin \alpha - \sin \beta = 2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right) \]
and
\[ \sin \alpha + \sin \beta = 2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right) \]

4. **Substitute these identities into the equation:**
Substituting the sine formulas into our equation gives:
\[ \frac{2 \cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)}{2 \sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)} = \frac{k - 1}{k + 1} \]

5. **Cancel the common factors:**
The factor of 2 cancels out:
\[ \frac{\cos\left(\frac{\alpha + \beta}{2}\right) \sin\left(\frac{\alpha - \beta}{2}\right)}{\sin\left(\frac{\alpha + \beta}{2}\right) \cos\left(\frac{\alpha - \beta}{2}\right)} = \frac{k - 1}{k + 1} \]

6. **Rewrite in terms of tangent and cotangent:**
This can be rewritten as:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) \cot\left(\frac{\alpha + \beta}{2}\right) = \frac{k - 1}{k + 1} \]

7. **Final result:**
Thus, we conclude that:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) \cot\left(\frac{\alpha + \beta}{2}\right) = \frac{k - 1}{k + 1} \]

### Final Answer:
\[ \tan\left(\frac{\alpha - \beta}{2}\right) \cot\left(\frac{\alpha + \beta}{2}\right) = \frac{k - 1}{k + 1} \]