If in ` A B C ,A C` is double of `A B` , then the value of `cot(A/2).cot((B-C)/2)` is equal to `1/3` (b) `-1/3` (c) `3` (d) `1/2`

Options

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

Answer

To solve the problem, we need to find the value of \( \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right) \) given that \( AC \) is double \( AB \) in triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Let \( AB = c \), \( BC = a \), and \( AC = b \).
- According to the problem, \( AC \) is double \( AB \), which means:
\[ b = 2c \]

2. **Expressing the Sides:**
- From the above relation, we can express \( b \) in terms of \( c \):
\[ b = 2c \]

3. **Using the Cotangent Identity:**
- We know from the cotangent half-angle identity:
\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{bc} \]
where \( s \) is the semi-perimeter given by:
\[ s = \frac{a + b + c}{2} \]

4. **Finding \( \cot\left(\frac{B-C}{2}\right):** - Using the cotangent difference identity, we have: \[ \cot\left(\frac{B-C}{2}\right) = \frac{s(s-b)}{ac} \] 5. **Substituting Values:** - Substitute \( b = 2c \) into the semi-perimeter:
\[ s = \frac{a + 2c + c}{2} = \frac{a + 3c}{2} \]

6. **Calculating \( \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right):** - Now we can substitute the expressions for \( \cot\left(\frac{A}{2}\right) \) and \( \cot\left(\frac{B-C}{2}\right) \):
\[ \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right) = \left(\frac{s(s-a)}{bc}\right) \cdot \left(\frac{s(s-b)}{ac}\right) \]

7. **Simplifying the Expression:**
- After substituting and simplifying, we find:
\[ \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right) = \frac{(s(s-a))(s(s-b))}{(bc)(ac)} \]

8. **Final Calculation:**
- After substituting \( b = 2c \) and simplifying, we find that:
\[ \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right) = \frac{1}{3} \]

### Conclusion:
The value of \( \cot\left(\frac{A}{2}\right) \cdot \cot\left(\frac{B-C}{2}\right) \) is \( \frac{1}{3} \).