Question

In a triangle \(ABC\), if \(a\neq b\), 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:

In triangle problems involving \(a\cos A\) and \(b\cos B\), use cosine rule identities to convert everything into side-based expressions.

Answer 1

The Correct Option is A

Step 1: Use cosine rule identities.
In a triangle \(ABC\), we have
\[ \cos A=\frac{b^2+c^2-a^2}{2bc} \] and
\[ \cos B=\frac{a^2+c^2-b^2}{2ac} \]
Now compute:
\[ a\cos A = a\left(\frac{b^2+c^2-a^2}{2bc}\right) \] \[ =\frac{a(b^2+c^2-a^2)}{2bc} \]
Similarly,
\[ b\cos B = b\left(\frac{a^2+c^2-b^2}{2ac}\right) \] \[ =\frac{b(a^2+c^2-b^2)}{2ac} \]

Step 2: Simplify numerator and denominator.
Using standard identities from triangle geometry,
\[ a\cos A-b\cos B = -\frac{(a-b)(a+b)\cos C}{c} \]
and
\[ a\cos B-b\cos A = \frac{(a-b)(a+b)}{c} \]
Therefore,
\[ \frac{a\cos A-b\cos B}{a\cos B-b\cos A} = -\cos C \]

Step 3: Evaluate the expression.
Hence,
\[ \frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos C \] \[ =-\cos C+\cos C \] \[ =0 \]

Step 4: Final conclusion.
Therefore,
\[ \boxed{0} \]