In a triangle ABC, r1-ra+r2-rb+r3-rc=

r1-ra=4 R sin2A22R2 sin A2cosA2=tanA2 Similarly r2-rb=tanB2,r3-rc=tanC2r1=s tanA2, r2=s tanB2, r3=s tanC2 r1+r2+r3s=tanA2+tanB2+tanC2

In a triangle ABC, r1-ra+r2-rb+r3-rc=

r1-ra=4 R sin2A22R2 sin A2cosA2=tanA2 Similarly r2-rb=tanB2,r3-rc=tanC2r1=s tanA2, r2=s tanB2, r3=s tanC2 r1+r2+r3s=tanA2+tanB2+tanC2

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In a triangle $ABC, \frac{r_{1} - r}{a} + \frac{r_{2 - r}}{b} + \frac{r_{3} - r}{c} =$

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$\frac{r_{1} + r_{2} + r_{3}}{2}$

$\frac{r_{1} + r_{2} + r_{3}}{s}$

$\frac{r_{1} + r_{2} + r_{3}}{3}$

$r_{1} + r_{2} + r_{3}$

answer is D.

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Detailed Solution

$\frac{r_{1} - r}{a} = \frac{4 R \sin^{2} \frac{A}{2}}{2 R (2 \sin \frac{A}{2} \cos \frac{A}{2})} = \tan \frac{A}{2}$ Similarly $\frac{r_{2} - r}{b} = \tan \frac{B}{2}, \frac{r_{3} - r}{c} = \tan \frac{C}{2}$

$r_{1} = s \tan \frac{A}{2}, r_{2} = s \tan \frac{B}{2}, r_{3} = s \tan \frac{C}{2} \frac{r_{1} + r_{2} + r_{3}}{s} = \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2}$