If in any Delta ABC,cot frac{A}{2}, cot frac{ | vedclass

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(B) Given that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$This implies $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}

Vedclass · Accepted Answer

$\cot \frac{A}{2} \cot \frac{C}{2} = 3$

Vedclass · Answer

(B) Given that $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$This implies $2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2}$.Using the identity $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$,we have $\cot \frac{A}{2} = \frac{s-a}{r}$,where $r$ is the inradius and $s$ is the semi-perimeter.Substituting this into the $A.P.$ condition: $2 \frac{s-b}{r} = \frac{s-a}{r} + \frac{s-c}{r}$.$2(s-b) = s-a + s-c \implies 2s - 2b = 2s - (a+c) \implies 2b = a+c$.This means $a, b, c$ are in $A.P.$For an equilateral triangle where $A=B=C=60^{\circ}$,$\cot 30^{\circ} = \sqrt{3}$.Then $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{3} 	imes \sqrt{3} = 3$. Thus,option $(b)$ is correct.

If in any $\Delta ABC$,$\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in $A.P.$,then:

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