In any triangle ABC,r^2 cot frac{A}{2} cot fr | vedclass

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(A) We know that in a triangle $ABC$,the cotangent half-angle formulas are $\cot \frac{A}{2} = \frac{s-a}{r}$,$\cot \frac{B}{2} = \frac{s-b}{r}$,and $

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$\Delta$

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(A) We know that in a triangle $ABC$,the cotangent half-angle formulas are $\cot \frac{A}{2} = \frac{s-a}{r}$,$\cot \frac{B}{2} = \frac{s-b}{r}$,and $\cot \frac{C}{2} = \frac{s-c}{r}$.Substituting these into the expression $r^2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$:$= r^2 \left( \frac{s-a}{r} \right) \left( \frac{s-b}{r} \right) \left( \frac{s-c}{r} \right)$$= r^2 \cdot \frac{(s-a)(s-b)(s-c)}{r^3}$$= \frac{(s-a)(s-b)(s-c)}{r}$Since the area of the triangle $\Delta = rs$,we have $r = \frac{\Delta}{s}$.Also,by Heron's formula,$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,so $\Delta^2 = s(s-a)(s-b)(s-c)$,which implies $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.Substituting these values:$= \frac{\Delta^2 / s}{\Delta / s} = \frac{\Delta^2}{s} \cdot \frac{s}{\Delta} = \Delta$.Thus,the correct option is $A$.

In any triangle $ABC$,$r^2 \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2} =$

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