299/ Potw

For a triangle ABC with the sides of a,b,c such that a + c = 2b prove that $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = 2 \cot(\frac{B}{2})$

Solution:
We have using law of cot for a triangle using standard notation s ,and area of triangle $\triangle$

$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$

$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$

$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

add (1) and (3) to get

 $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = \dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}$
$= \dfrac{s(s-a)+s(s-c)}{\triangle}$
$= \dfrac{s(s-a+s-c)}{\triangle}$
$= \dfrac{s(2s-(a+c))}{\triangle}$
$= \dfrac{s(2s-2b)}{\triangle}$ (from given conditon)
$= 2\dfrac{s(s-b)}{\triangle}$
$=2\cot(\frac{B}{2})$ (using (2))

proved