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

In a triangle ABC with usual notations, $\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}=$
  1. $\frac{1}{b}$
  2. $\frac{2}{b}$
  3. $\frac{-1}{b}$
  4. $\frac{-2}{b}$

Solution

$\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}$ $=\frac{b \cos A-b \cos C+a \cos B-c \cos B}{b(a-c)}$ $=\frac{(a \cos B+b \cos A)-(b \cos C+c \cos B)}{b(a-c)}$ $=\frac{c-a}{b(a-c)}=\frac{-1}{b}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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