We know thatr=4Rsin⁡A2sin⁡B2sin⁡C2r1=4Rsin⁡A2cos⁡B2cos⁡C2r2=4Rcos⁡A2sin⁡B2cos⁡C2r3=4Rcos⁡A2cos⁡B2sin⁡C2 Now r2+r3=4RcosA2(sinB2cosC2+cosB2sinC2) =4Rcos⁡A2sin⁡(B+C2)→(1) Now r1-r =4RsinA2 (cosB2cosC2-sinB2sinC2) =4RsinA2cos(B+C2)→(2) (1)-(2)⇒r2+r3 -(r1-r)=4Rcos⁡A2sin⁡(B+C2)-4RsinA2cos(B+C2) =4Rsin(B+C-A2) =4Rsin(90∘ -A) = 4RcosA

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In $ΔABC, r + r_{2} + r_{3} - r_{1} =$

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4 R cos A

4 R cos B

4 R cos C

answer is A.

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Detailed Solution

$\begin{array}{l} W e k n o w t h a t \\ r = 4 Rsin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ r_{1} = 4 Rsin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ r_{2} = 4 Rcos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} \end{array}$
$r_{3} = 4 Rcos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
$Now r_{2} + r_{3} = 4 R \cos \frac{A}{2} (\sin \frac{B}{2} \cos \frac{C}{2} + \cos \frac{B}{2} \sin \frac{C}{2}) = 4 R \cos \frac{A}{2} \sin (\frac{B + C}{2}) \to (1) N o w r_{1} - r_{} = 4 R \sin \frac{A}{2} (\cos \frac{B}{2} \cos \frac{C}{2} - \sin \frac{B}{2} \sin \frac{C}{2}) = 4 R \sin \frac{A}{2} \cos (\frac{B + C}{2}) \to (2) (1) - (2) \Rightarrow r_{2} + r_{3} - (r_{1} - r) = 4 R \cos \frac{A}{2} \sin (\frac{B + C}{2}) - 4 R \sin \frac{A}{2} \cos (\frac{B + C}{2}) = 4 R \sin (\frac{B + C - A}{2}) = 4 R \sin (90^{\circ} - A) = 4 R \cos A$