If the sides of a triangle are in A. P., then the cotangent of its half the angles will be in  [MP PET 1993]

Let \[\cot \frac{A}{2},\cot \frac{B}{2}\]and \[\cot \frac{C}{2}\]be in A.P. , then \[2\cot \frac{B}{2}=\cot \frac{C}{2}+\cot \frac{A}{2}\] Þ \[2\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}\]\[+\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\] R.H.S = \[\sqrt{\frac{s}{(s-b)}}\LEFT( \sqrt{\frac{(s-c)}{(s-a)}}+\sqrt{\frac{(s-a)}{(s-c)}} \right)\] \[=\sqrt{\frac{s}{s-b}}\left( \frac{s-c+s-a}{\sqrt{(s-a)(s-c)}} \right)\]\[=\sqrt{\frac{s}{s-b}}\left( \frac{2s-a-c}{\sqrt{(s-a)(s-c)}} \right)\]    \[=2\sqrt{\frac{s}{(s-b)}}\sqrt{\frac{{{(s-b)}^{2}}}{(s-a)(s-c)}}\], \[\{\because a+c=2b\], \[a+b+c=2s\] i.e., \[2(s-b)=2s-a-c\}\] \[=2\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}\]=L.H.S. Note: STUDENTS should remember this question as a fact.