In triangle ABC,the expression frac{16 R s De | vedclass

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(D) We know that in $	riangle ABC$,the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Also

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$4 r_1 r_2 r_3$

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(D) We know that in $	riangle ABC$,the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Also,$\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,$\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$,and $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$.Substituting these into the expression:$E = \frac{16 R s \Delta}{s-c} \cdot \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-a)(s-c)}{ac}} \cdot \sqrt{\frac{s(s-c)}{ab}}$$E = \frac{16 R s \Delta}{s-c} \cdot \frac{(s-c) \sqrt{s(s-a)(s-b)(s-c)}}{abc}$Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ and $abc = 4R\Delta$,we have:$E = \frac{16 R s \Delta}{s-c} \cdot \frac{(s-c) \Delta}{4R\Delta} = \frac{16 R s \Delta}{4R} = 4s\Delta$.Now,$r_1 r_2 r_3 = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c} = \frac{\Delta^3}{(s-a)(s-b)(s-c)}$.Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$.Thus,$r_1 r_2 r_3 = \frac{\Delta^3}{\Delta^2/s} = s\Delta$.Therefore,$4s\Delta = 4 r_1 r_2 r_3$.

In $\triangle ABC$,the expression $\frac{16 R s \Delta \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}}{s-c}$ is equal to

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