In `Delta ABC, "cot"(A)/(2) + "cot" (B)/(2) + "cot" (C)/(2)` is equal to

Options

A. `(Delta)/(r^(2))`
B. `((a + b + c)^(2))/(abc) 2R`
C. `(Delta )/(r)`
D. `(Delta)/(Rr)`

Explanation

To solve the problem, we need to find the value of \(\frac{\cot(A/2)}{2} + \frac{\cot(B/2)}{2} + \frac{\cot(C/2)}{2}\) in triangle \(ABC\).

### Step-by-Step Solution:

1. **Write the Given Expression**:
\[ \frac{\cot(A/2)}{2} + \frac{\cot(B/2)}{2} + \frac{\cot(C/2)}{2} \]

2. **Use the Formula for Cotangent of Half Angles**:
The formulas for \(\cot(A/2)\), \(\cot(B/2)\), and \(\cot(C/2)\) in terms of the sides of the triangle are:
\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{\Delta}, \quad \cot\left(\frac{B}{2}\right) = \frac{s(s-b)}{\Delta}, \quad \cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{\Delta} \]
where \(s\) is the semi-perimeter \(s = \frac{a+b+c}{2}\) and \(\Delta\) is the area of the triangle.

3. **Substitute the Formulas into the Expression**:
Substitute the formulas into the expression:
\[ \frac{1}{2} \left( \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta} \right) \]

4. **Combine the Terms**:
Factor out the common terms:
\[ = \frac{1}{2} \cdot \frac{s}{\Delta} \left( (s-a) + (s-b) + (s-c) \right) \]

5. **Simplify the Expression Inside the Parentheses**:
Simplifying the expression inside the parentheses:
\[ (s-a) + (s-b) + (s-c) = 3s - (a+b+c) \]
Since \(a + b + c = 2s\), we have:
\[ 3s - 2s = s \]

6. **Final Expression**:
Substitute back into the expression:
\[ = \frac{1}{2} \cdot \frac{s}{\Delta} \cdot s = \frac{s^2}{2\Delta} \]

7. **Relate to the Inradius**:
We know that the inradius \(r\) is given by:
\[ r = \frac{\Delta}{s} \]
Therefore, \(\Delta = sr\). Substitute this into the expression:
\[ = \frac{s^2}{2(sr)} = \frac{s}{2r} \]

### Conclusion:
Thus, the final result is:
\[ \frac{s}{2r} \]