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

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(A) We know that in a triangle $ABC$, the exradii are given by $r_2 = s 	an \left(\frac{B}{2}ight)$ and $r_3 = s 	an \left(\frac{C}{2}ight)$.Als

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$4R \cot \left(\frac{A}{2}\right)$

Vedclass · Answer

(A) We know that in a triangle $ABC$, the exradii are given by $r_2 = s 	an \left(\frac{B}{2}ight)$ and $r_3 = s 	an \left(\frac{C}{2}ight)$.Also, $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Using the identity $r_2 + r_3 = \frac{\Delta}{s-b} + \frac{\Delta}{s-c} = \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} ight) = \Delta \left( \frac{2s-b-c}{(s-b)(s-c)} ight) = \Delta \left( \frac{a}{(s-b)(s-c)} ight)$.Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, we have $\frac{\Delta}{(s-b)(s-c)} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} = \cot \left(\frac{A}{2}ight)$.Thus, $r_2 + r_3 = a \cot \left(\frac{A}{2}ight)$.Using the sine rule, $a = 2R \sin A = 2R \cdot 2 \sin \left(\frac{A}{2}ight) \cos \left(\frac{A}{2}ight) = 4R \sin \left(\frac{A}{2}ight) \cos \left(\frac{A}{2}ight)$.Substituting this into the expression:$(r_2 + r_3) \operatorname{cosec}^2 \left(\frac{A}{2}ight) = 4R \sin \left(\frac{A}{2}ight) \cos \left(\frac{A}{2}ight) \cdot \frac{1}{\sin^2 \left(\frac{A}{2}ight)} = 4R \cot \left(\frac{A}{2}ight)$.

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

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