if cosA=m cos B and `cot((A+B)/(2))=lamda`. `tan((B-A)/(2))` then `lamda` is equal to

Options

A. `(m)/(m-1)`
B. `(m+1)/(m)`
C. `(m+1)/(m-1)`
D. none of these

Explanation

To solve the problem, we start with the given equation:

1. **Given**: \( \cos A = m \cos B \)

2. **Rearranging**: We can express this as:
\[ \frac{\cos A}{\cos B} = m \]

3. **Using the Cosine Addition Formula**: We can apply the formula for the sum and difference of cosines:
\[ \frac{\cos A + \cos B}{\cos A - \cos B} = \frac{m + 1}{m - 1} \]

4. **Applying the Cosine Addition and Subtraction Formulas**:
\[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]

5. **Substituting these into the equation**:
\[ \frac{2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)}{-2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)} = \frac{m + 1}{m - 1} \]

6. **Simplifying**:
\[ -\frac{\cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)}{\sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)} = \frac{m + 1}{m - 1} \]

7. **Using the cotangent identity**:
\[ -\cot\left(\frac{A + B}{2}\right) \cdot \frac{\cos\left(\frac{A - B}{2}\right)}{\sin\left(\frac{A - B}{2}\right)} = \frac{m + 1}{m - 1} \]

8. **Recognizing that**:
\[ \tan\left(\frac{B - A}{2}\right) = \frac{\sin\left(\frac{B - A}{2}\right)}{\cos\left(\frac{B - A}{2}\right)} \]

9. **Setting up the final equation**:
\[ \lambda = \frac{m + 1}{m - 1} \]

Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{m + 1}{m - 1} \]