If triangle ABC is right-angled at A, then r_ | vedclass

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(A) In a triangle $ABC$, the exradii are given by $r_1 = s 	an(A/2)$, $r_2 = s 	an(B/2)$, and $r_3 = s 	an(C/2)$, and the inradius is $r = (s-a)

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$r_1-r$

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(A) In a triangle $ABC$, the exradii are given by $r_1 = s 	an(A/2)$, $r_2 = s 	an(B/2)$, and $r_3 = s 	an(C/2)$, and the inradius is $r = (s-a) 	an(A/2)$.Given $\angle A = 90^{\circ}$, we have $A/2 = 45^{\circ}$.Thus, $r_1 = s 	an(45^{\circ}) = s$.Also, $r = (s-a) 	an(45^{\circ}) = s-a$.Therefore, $r_1 - r = s - (s-a) = a$.In a right-angled triangle, the hypotenuse $a = 2R$, where $R$ is the circumradius.Now, $r_2 + r_3 = s(	an(B/2) + 	an(C/2))$.Since $B+C = 90^{\circ}$, $C/2 = 45^{\circ} - B/2$, so $	an(C/2) = 	an(45^{\circ} - B/2) = \frac{1-	an(B/2)}{1+	an(B/2)}$.Substituting this, $r_2 + r_3 = s \left( 	an(B/2) + \frac{1-	an(B/2)}{1+	an(B/2)} ight) = s \left( \frac{	an(B/2) + 	an^2(B/2) + 1 - 	an(B/2)}{1+	an(B/2)} ight) = s \frac{1+	an^2(B/2)}{1+	an(B/2)}$.Using the identity $r_2+r_3 = 4R \cos(A/2) \cos(B/2) \cos(C/2) / \sin(A/2)$ is complex, but we know $r_1+r_2+r_3-r = 4R$. For $A=90^{\circ}$, $r_1 = s$. Since $r_1 = 4R \cos(A/2) \sin(B/2) \sin(C/2) / \cos(A/2) = 4R \sin(B/2) \sin(C/2)$, and $r_2+r_3 = 4R \cos(A/2) \cos(B/2) \cos(C/2) / \sin(A/2)$, for $A=90^{\circ}$, $r_2+r_3 = 2R \sqrt{2} \cos(B/2) \cos(C/2) 	imes 2 = 2R(1+\sin A) = 2R(1+1) = 4R$ is incorrect. Actually, $r_1-r = 4R \sin^2(A/2) = 4R \sin^2(45^{\circ}) = 2R$. Since $r_2+r_3 = 2R$, the correct relation is $r_2+r_3 = r_1-r$.

If $\triangle ABC$ is right-angled at $A$, then $r_2+r_3$ is equal to

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