In a `Delta ABC b= sqrt(3)+1, c=sqrt(3)-1 , angle A = 60 ^@` then the value of tan `(B-C)/(2)` is :-

Options

A. 2
B. `1//2`
C. 1
D. 3

Answer

To solve the problem, we will use Napier's analogy to find the value of \( \tan \left( \frac{B - C}{2} \right) \).

### Step-by-step Solution:

1. **Identify Given Values**:
- \( b = \sqrt{3} + 1 \)
- \( c = \sqrt{3} - 1 \)
- \( A = 60^\circ \)

2. **Use Napier's Analogy**:
According to Napier's analogy:
\[ \tan \left( \frac{B - C}{2} \right) = \frac{b - c}{b + c} \cdot \cot \left( \frac{A}{2} \right) \]

3. **Calculate \( b - c \)**:
\[ b - c = (\sqrt{3} + 1) - (\sqrt{3} - 1) = 1 + 1 = 2 \]

4. **Calculate \( b + c \)**:
\[ b + c = (\sqrt{3} + 1) + (\sqrt{3} - 1) = 2\sqrt{3} \]

5. **Calculate \( \cot \left( \frac{A}{2} \right) \)**:
Since \( A = 60^\circ \):
\[ \frac{A}{2} = 30^\circ \quad \text{and} \quad \cot(30^\circ) = \sqrt{3} \]

6. **Substitute Values into Napier's Formula**:
\[ \tan \left( \frac{B - C}{2} \right) = \frac{2}{2\sqrt{3}} \cdot \sqrt{3} \]

7. **Simplify the Expression**:
\[ \tan \left( \frac{B - C}{2} \right) = \frac{2\sqrt{3}}{2\sqrt{3}} = 1 \]

### Final Answer:
Thus, the value of \( \tan \left( \frac{B - C}{2} \right) \) is \( 1 \).

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