In triangle ABC, (r_2 + r_3) cot left(frac{B+ | vedclass

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(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Also,$\cot \left(\frac{B+C}{2}ight) = 	an \left(\frac{A}{2}ight) = \sq

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$a$

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(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Also,$\cot \left(\frac{B+C}{2}ight) = 	an \left(\frac{A}{2}ight) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.Substituting these values:$(r_2 + r_3) \cot \left(\frac{B+C}{2}ight) = \left( \frac{\Delta}{s-b} + \frac{\Delta}{s-c} ight) 	an \left(\frac{A}{2}ight)$$= \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} ight) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$= \Delta \left( \frac{a}{(s-b)(s-c)} ight) \frac{\sqrt{(s-b)(s-c)}}{\sqrt{s(s-a)}}$$= \frac{\Delta \cdot a}{\sqrt{s(s-a)(s-b)(s-c)}}$Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,the expression simplifies to:$= \frac{\Delta \cdot a}{\Delta} = a$.Thus,option $B$ is correct.

In $\triangle ABC, (r_2 + r_3) \cot \left(\frac{B+C}{2}\right) = $

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