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In a triangle ABC with usual notations, $\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}=$
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Verified Answer
The correct answer is:
$\frac{-1}{b}$
$\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}$
$=\frac{b \cos A-b \cos C+a \cos B-c \cos B}{b(a-c)}$
$=\frac{(a \cos B+b \cos A)-(b \cos C+c \cos B)}{b(a-c)}$
$=\frac{c-a}{b(a-c)}=\frac{-1}{b}$
$=\frac{b \cos A-b \cos C+a \cos B-c \cos B}{b(a-c)}$
$=\frac{(a \cos B+b \cos A)-(b \cos C+c \cos B)}{b(a-c)}$
$=\frac{c-a}{b(a-c)}=\frac{-1}{b}$
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