In `Delta ABC`, If `a =4`, `b=3` and `COS(A-B) =3/4,` then

Options

A. measure of `angle A` is `pi/2`
B. measuere of `angle B is pi/2`
C. `cot""C/2=sqrt7`
D. circumradius of `Delta ABC is (2)/(7 ^(1//14))`

Answer

To solve the problem, we need to find the value of \( \cot \frac{C}{2} \) in triangle \( ABC \) given the sides \( a = 4 \), \( b = 3 \), and \( \cos(A - B) = \frac{3}{4} \).

### Step-by-step Solution:

1. **Use the formula for \( \tan \frac{A - B}{2} \)**:
\[ \tan \frac{A - B}{2} = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} \]
Substituting \( \cos(A - B) = \frac{3}{4} \):
\[ \tan \frac{A - B}{2} = \sqrt{\frac{1 - \frac{3}{4}}{1 + \frac{3}{4}} = \sqrt{\frac{\frac{1}{4}}{\frac{7}{4}} = \sqrt{\frac{1}{7}} = \frac{1}{\sqrt{7}}} \]

2. **Relate \( \tan \frac{A - B}{2} \) to \( \cot \frac{C}{2} \)**:
The relationship between \( \tan \frac{A - B}{2} \) and \( \cot \frac{C}{2} \) is given by:
\[ \tan \frac{A - B}{2} = \frac{a - b}{a + b} \cdot \cot \frac{C}{2} \]
Substituting \( a = 4 \) and \( b = 3 \):
\[ \tan \frac{A - B}{2} = \frac{4 - 3}{4 + 3} \cdot \cot \frac{C}{2} = \frac{1}{7} \cdot \cot \frac{C}{2} \]

3. **Set the two expressions for \( \tan \frac{A - B}{2} \) equal**:
\[ \frac{1}{\sqrt{7}} = \frac{1}{7} \cdot \cot \frac{C}{2} \]

4. **Solve for \( \cot \frac{C}{2} \)**:
Multiply both sides by 7:
\[ 7 \cdot \frac{1}{\sqrt{7}} = \cot \frac{C}{2} \]
Simplifying gives:
\[ \cot \frac{C}{2} = \frac{7}{\sqrt{7}} = \sqrt{7} \]

Thus, the value of \( \cot \frac{C}{2} \) is \( \sqrt{7} \).

### Final Answer:
\[ \cot \frac{C}{2} = \sqrt{7} \]