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

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