In triangle ABC, (r_2+r_3) operatorname{cosec | vedclass

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(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Also,$\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$.Thus,$(r_2+r_3) \operator

Vedclass · Accepted Answer

$4 R \cot^2 \frac{A}{2}$

Vedclass · Answer

(B) We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Also,$\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$.Thus,$(r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} = \left(\frac{\Delta}{s-b} + \frac{\Delta}{s-c}ight) 	imes \frac{bc}{(s-b)(s-c)}$.$= \Delta \left(\frac{s-c+s-b}{(s-b)(s-c)}ight) 	imes \frac{bc}{(s-b)(s-c)} = \Delta \left(\frac{a}{(s-b)(s-c)}ight) 	imes \frac{bc}{(s-b)(s-c)}$.$= \frac{\Delta abc}{(s-b)^2(s-c)^2} = \frac{4R \Delta^2}{(s-b)^2(s-c)^2}$.$= 4R \left(\frac{\Delta}{(s-b)(s-c)}ight)^2 = 4R \left(\cot \frac{A}{2}ight)^2 = 4R \cot^2 \frac{A}{2}$.

In $\triangle ABC, (r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} =$

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