In a triangle `ABC` if `cot(A/2)cot(B/2)=c, cot(B/2)cot(C/2)=a ` and `cot(C/2)cot(A/2)=b` then `1/(s-a)+1/(s-b)+1/(s-c)=`

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem step by step, we will follow the given conditions and derive the required expression.

### Step 1: Understand the Given Conditions
We are given:
1. \( \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) = c \)
2. \( \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) = a \)
3. \( \cot\left(\frac{C}{2}\right) \cot\left(\frac{A}{2}\right) = b \)

### Step 2: Express \( s \)
The semi-perimeter \( s \) of triangle \( ABC \) is given by:
\[ s = \frac{a + b + c}{2} \]

### Step 3: Rewrite the Cotangent Relationships
Using the relationships given, we can express each cotangent in terms of \( s \):
1. From \( \cot\left(\frac{A}{2}\right) \cot\left(\frac{B}{2}\right) = c \):
\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{(s-b)(s-c)} \]
Therefore,
\[ \frac{s}{s-c} = c \quad \text{(Equation 1)} \]

2. From \( \cot\left(\frac{B}{2}\right) \cot\left(\frac{C}{2}\right) = a \):
\[ \cot\left(\frac{B}{2}\right) = \frac{s(s-b)}{(s-a)(s-c)} \]
Therefore,
\[ \frac{s}{s-a} = a \quad \text{(Equation 2)} \]

3. From \( \cot\left(\frac{C}{2}\right) \cot\left(\frac{A}{2}\right) = b \):
\[ \cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{(s-a)(s-b)} \]
Therefore,
\[ \frac{s}{s-b} = b \quad \text{(Equation 3)} \]

### Step 4: Find the Required Expression
We need to find:
\[ \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} \]

Using Equations 1, 2, and 3, we can express \( \frac{1}{s-a}, \frac{1}{s-b}, \frac{1}{s-c} \):
- From Equation 1: \( \frac{1}{s-a} = \frac{1}{s} \cdot \frac{1}{c} \)
- From Equation 2: \( \frac{1}{s-b} = \frac{1}{s} \cdot \frac{1}{a} \)
- From Equation 3: \( \frac{1}{s-c} = \frac{1}{s} \cdot \frac{1}{b} \)

### Step 5: Combine the Expressions
Combining these gives:
\[ \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{1}{s} \left( \frac{1}{c} + \frac{1}{a} + \frac{1}{b} \right) \]

### Step 6: Simplify the Expression
Now, we know that:
\[ \frac{1}{c} + \frac{1}{a} + \frac{1}{b} = \frac{a+b+c}{abc} \]
Thus,
\[ \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{a+b+c}{s \cdot abc} \]

### Step 7: Substitute \( s \)
Since \( s = \frac{a+b+c}{2} \), we can substitute this into our expression:
\[ \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{2}{abc} \]

### Final Result
Thus, the final answer is:
\[ \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = 2 \]