MHT CET · Maths · Properties of Triangles
In a triangle ABC with usual notations, \(\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}=\)
- A \(\frac{1}{b}\)
- B \(\frac{2}{b}\)
- C \(\frac{-1}{b}\)
- D \(\frac{-2}{b}\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{b}\)
Step-by-step Solution
Detailed explanation
\(\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}\)
\(=\frac{b \cos A-b \cos C+a \cos B-c \cos B}{b(a-c)}\)
\(=\frac{(a \cos B+b \cos A)-(b \cos C+c \cos B)}{b(a-c)}\)
\(=\frac{c-a}{b(a-c)}=\frac{-1}{b}\)
\(=\frac{b \cos A-b \cos C+a \cos B-c \cos B}{b(a-c)}\)
\(=\frac{(a \cos B+b \cos A)-(b \cos C+c \cos B)}{b(a-c)}\)
\(=\frac{c-a}{b(a-c)}=\frac{-1}{b}\)
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