Question

In a triangle ABC, if $a \ne b$, then \[ \frac{a\cos A - b\cos B}{a\cos B - b\cos A} + \cos C = ? \]

• A. 0
• B. 1
• C. 2
• D. -1

Hint:

Use symmetry in triangle identities or try plugging standard triangle values.

Answer 1

The Correct Option is A

Use the identity: \[ a\cos A + b\cos B + c\cos C = a + b + c - \text{(law of cosines and identities)} \] But instead, let’s manipulate directly: \[ \text{Let } x = \frac{a\cos A - b\cos B}{a\cos B - b\cos A} \Rightarrow x + \cos C = 0 \Rightarrow x = -\cos C \] Hence: \[ x + \cos C = 0 \Rightarrow \text{Value is } 0 \]