In triangle ABC, find the value of frac{1+cos | vedclass

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(C) We know that for a triangle $ABC$ with exradii $r_1, r_2, r_3$ and circumradius $R$: $r_1+r_2 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}

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$\frac{3}{2R}$

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(C) We know that for a triangle $ABC$ with exradii $r_1, r_2, r_3$ and circumradius $R$: $r_1+r_2 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2}$ $= 4R \cos \frac{C}{2} [\sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2}]$ $= 4R \cos \frac{C}{2} \sin(\frac{A+B}{2}) = 4R \cos \frac{C}{2} \cos \frac{C}{2} = 2R(2 \cos^2 \frac{C}{2}) = 2R(1+\cos C)$. Thus, $\frac{1+\cos C}{r_1+r_2} = \frac{1}{2R}$. Similarly, $\frac{1+\cos A}{r_2+r_3} = \frac{1}{2R}$ and $\frac{1+\cos B}{r_1+r_3} = \frac{1}{2R}$. Adding these, we get $\frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R}$.

In $\triangle ABC$, find the value of $\frac{1+\cos C}{r_1+r_2}+\frac{1+\cos A}{r_2+r_3}+\frac{1+\cos B}{r_1+r_3}$.

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