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 the expression:

\[ \frac{\cot(A/2)}{2} + \frac{\cot(B/2)}{2} + \frac{\cot(C/2)}{2} \]

### Step 1: Use the Half-Angle Cotangent Formula
We know that:

\[ \cot\left(\frac{A}{2}\right) = \frac{s(s-a)}{\Delta}, \quad \cot\left(\frac{B}{2}\right) = \frac{s(s-b)}{\Delta}, \quad \cot\left(\frac{C}{2}\right) = \frac{s(s-c)}{\Delta} \]

where \( s \) is the semi-perimeter of the triangle, \( s = \frac{A + B + C}{2} \), and \( \Delta \) is the area of triangle ABC.

### Step 2: Substitute the Values
Substituting these values into our expression gives:

\[ \frac{1}{2} \left( \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta} \right) \]

### Step 3: Factor Out Common Terms
We can factor out \( \frac{s}{\Delta} \):

\[ = \frac{s}{2\Delta} \left( (s-a) + (s-b) + (s-c) \right) \]

### Step 4: Simplify the Expression Inside the Parentheses
Now, simplifying the expression inside the parentheses:

\[ (s-a) + (s-b) + (s-c) = 3s - (a + b + c) \]

Since \( a + b + c = 2s \), we have:

\[ 3s - 2s = s \]

### Step 5: Substitute Back
Substituting this back into our expression gives:

\[ = \frac{s}{2\Delta} \cdot s = \frac{s^2}{2\Delta} \]

### Step 6: Final Expression
Thus, we have:

\[ \frac{s^2}{2\Delta} \]

### Conclusion
The final expression for \( \frac{\cot(A/2)}{2} + \frac{\cot(B/2)}{2} + \frac{\cot(C/2)}{2} \) is:

\[ \frac{s^2}{2\Delta} \]

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