Question

In a triangle ABC with usual notations, \(\cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=\)

• A. \(\frac{\mathrm{s}^{2}}{\Delta}\), where \(\Delta\) is the area of the triangle ABC .
• B. \(\frac{s}{\Delta}\), where \(\Delta\) is the area of the triangle ABC .
• C. \(\frac{\Delta}{\mathrm{s}}\), where \(\Delta\) is the area of the triangle ABC .
• D. \(\Delta\), where \(\Delta\) is the area of the triangle ABC .

Hint:

Use cot(A/2) = s(s - a)/Δ for each angle and add.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a triangle, \(\cot\dfrac{A}{2} = \dfrac{s(s-a)}{\Delta}\), and similarly for \(B\) and \(C\).

Step 2: Add the three
\(\cot\dfrac{A}{2} + \cot\dfrac{B}{2} + \cot\dfrac{C}{2} = \dfrac{s\left[(s-a) + (s-b) + (s-c)\right]}{\Delta}\).

Step 3: Simplify
\((s-a) + (s-b) + (s-c) = 3s - 2s = s\), so the sum is \(\dfrac{s^2}{\Delta}\).

Final Answer:
The sum equals \(s^2/\Delta\). \[ \boxed{\text{(A) }\dfrac{s^{2}}{\Delta}} \]