Multiple choice

In $\Delta$ $ABC$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively. Then $\dfrac { \cot { \dfrac { A }{ 2 } } +\cot { \dfrac { B }{ 2 } } +\cot { \dfrac { C }{ 2 } } }{ \cot { A } +\cot { B } +\cot { C } } =?$

  1. $\displaystyle \frac { 4{ s }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } $
  2. $\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 2s } $
  3. $\displaystyle \frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ 3s } $

  4. None of these
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A Correct answer
Explanation

Using standard trigonometric identities in a triangle, the sum of cot(A/2) is s/r and the sum of cot(A) is (a^2+b^2+c^2)/(4*Area). The ratio simplifies to the given expression involving s and side lengths.

AI explanation

Using the half-angle identity, cot of A divided by 2 equals the semi-perimeter s minus side a divided by the inradius r, so the sum of the cotangents of the half angles equals (3s minus the sum of the sides) divided by r. Since 3s minus the sum of the sides equals s, this numerator simplifies to s over r. For the denominator, using the identity cot A plus cot B plus cot C equals (a squared plus b squared plus c squared) divided by 4 times the area, and knowing the area equals r times s, the denominator becomes (a squared plus b squared plus c squared) divided by 4 times r times s. Dividing the numerator by the denominator gives 4 times s squared divided by (a squared plus b squared plus c squared).

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