Question

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

• A. \( \dfrac{1}{b} \)
• B. \( \dfrac{2}{b} \)
• C. \( -\dfrac{1}{b} \)
• D. \( -\dfrac{2}{b} \)

Hint:

In triangle identities involving sides and cosines, converting everything using the cosine rule often leads to quick cancellation.

Answer 1

The Correct Option is C

Step 1: Use cosine rule relations.
In triangle \(ABC\), \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \] Step 2: Compute \(\cos A - \cos C\).
\[ \cos A - \cos C = \frac{(b^2 + c^2 - a^2)a - (a^2 + b^2 - c^2)c}{2abc} = \frac{(a-c)(b^2 - (a-c)^2)}{2abc} \] Hence, \[ \frac{\cos A - \cos C}{a-c} = \frac{b^2 - (a-c)^2}{2abc} \] Step 3: Add \(\dfrac{\cos B}{b}\).
Using \(\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}\), we get \[ \frac{\cos B}{b} = \frac{a^2 + c^2 - b^2}{2abc} \] Step 4: Simplify.
Adding both terms, \[ \frac{b^2 - (a-c)^2 + a^2 + c^2 - b^2}{2abc} = \frac{-2ac}{2abc} = -\frac{1}{b} \]