In a triangle ABC,the expression frac{cos C+c | vedclass

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(B) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the first term: $\frac{\cos C+\cos A}{c+a} = \fra

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$\frac{1}{b}$

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(B) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these into the first term: $\frac{\cos C+\cos A}{c+a} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(\sin C+\sin A)} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(2 \sin \frac{C+A}{2} \cos \frac{C-A}{2})} = \frac{1}{2R} \cot \frac{C+A}{2}$. Since $A+B+C = \pi$,$\frac{C+A}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cot \frac{C+A}{2} = 	an \frac{B}{2}$. Thus,the first term is $\frac{1}{2R} 	an \frac{B}{2}$. For the second term: $\frac{\cos B}{b} = \frac{\cos B}{2R \sin B} = \frac{1}{2R} \cot B = \frac{1}{2R} \left( \frac{1-	an^2 \frac{B}{2}}{2 	an \frac{B}{2}} ight)$. Adding both terms: $\frac{1}{2R} \left( 	an \frac{B}{2} + \frac{1-	an^2 \frac{B}{2}}{2 	an \frac{B}{2}} ight) = \frac{1}{2R} \left( \frac{2 	an^2 \frac{B}{2} + 1 - 	an^2 \frac{B}{2}}{2 	an \frac{B}{2}} ight) = \frac{1}{2R} \left( \frac{1+	an^2 \frac{B}{2}}{2 	an \frac{B}{2}} ight) = \frac{1}{2R \sin B} = \frac{1}{b}$.

In a $\triangle ABC$,the expression $\frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b}$ is equal to

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