Q69 / 150 · MHT CET 2025 (23 Apr Shift 1) · Mathematics

In a triangle ABC with usual notations. $\cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=$

  • Correct answer
    A
    $\frac{s^2}{\Delta}$, where $\Delta$ is the area of the triangle $A B C$.
  • B
    $\frac{s}{\Delta}$, where $\Delta$ is the area of the triangle $A B C$.
  • C
    $\frac{\triangle}{s}$, where $\triangle$ is the area of the triangle $A B C$.
  • D
    $\Delta$, where $\Delta$ is the area of the triangle $A B C$.

Solution

Using the half-angle cotangent formulas for triangle ABC, with semi-perimeter $s = \frac{a+b+c}{2}$ and area $\Delta$:

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

The sum becomes:

$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \frac{s(s-a) + s(s-b) + s(s-c)}{\Delta} = \frac{s[(s-a) + (s-b) + (s-c)]}{\Delta}$

Simplifying the numerator:

$s[(s-a) + (s-b) + (s-c)] = s[3s - (a+b+c)] = s[3s - 2s] = s^2$

Thus:

$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \frac{s^2}{\Delta}$

This corresponds to option A.

Hence, the correct answer is option: 'A'.

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