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(C) Let the sides of the triangle be $a, b, c$ such that they are in $A.P.$,which implies $a + c = 2b$. We need to check if $\cot \frac{A}{2}, \cot \f

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$A.P.$

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(C) Let the sides of the triangle be $a, b, c$ such that they are in $A.P.$,which implies $a + c = 2b$. We need to check if $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$Recall the formula $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$. For $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ to be in $A.P.$,we must satisfy $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2}$. Substituting the formulas: $2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} + \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$. Dividing by $\sqrt{s}$,we get $2 \sqrt{\frac{s-b}{(s-a)(s-c)}} = \frac{s-c + s-a}{\sqrt{(s-a)(s-b)(s-c)}} \cdot \sqrt{s-b} = \frac{2s-a-c}{\sqrt{(s-a)(s-b)(s-c)}}$. Since $a+c=2b$ and $a+b+c=2s$,we have $2s-a-c = 2b$. Also $2s-a-c = 2(s-b)$. The expression simplifies to $2 \sqrt{\frac{s-b}{(s-a)(s-c)}} = \frac{2(s-b)}{\sqrt{(s-a)(s-b)(s-c)}} = 2 \sqrt{\frac{s-b}{(s-a)(s-c)}}$. Thus,the condition holds,and the cotangents of the half-angles are in $A.P.$

If the sides of a triangle are in $A.P.$,then the cotangents of its half-angles will be in

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