In a triangle ABC with usual notations,frac{c | vedclass

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(C) We are given the expression $\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}$.By taking the common denominator,we get $\frac{b(\cos A-\cos C) + (a-c)\c

Vedclass · Accepted Answer

$\frac{-1}{b}$

Vedclass · Answer

(C) We are given the expression $\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}$.By taking the common denominator,we get $\frac{b(\cos A-\cos C) + (a-c)\cos B}{b(a-c)}$.Expanding the numerator,we have $\frac{b \cos A - b \cos C + a \cos B - c \cos B}{b(a-c)}$.Rearranging the terms,we get $\frac{(a \cos B + b \cos A) - (b \cos C + c \cos B)}{b(a-c)}$.Using the projection formula $c = a \cos B + b \cos A$ and $a = b \cos C + c \cos B$,the expression becomes $\frac{c - a}{b(a-c)}$.Since $c - a = -(a - c)$,the expression simplifies to $\frac{-(a - c)}{b(a - c)} = \frac{-1}{b}$.

In a triangle $ABC$ with usual notations,$\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}=$

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