A) \[\frac{1}{a}\]
B) \[\frac{1}{b}\]
C) \[\frac{1}{c}\]
D) \[\frac{c+a}{b}\]
Correct Answer: B
Solution :
\[\frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b}\] \[=\frac{(b\cos C+c\cos B)+(b\cos A+a\cos B)}{b(c+a)}\] \[=\frac{a+c}{b(c+a)}\] , (using Projection formulae) \[=\frac{1}{b}\].
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