Question

In any $\triangle ABC$, with usual notations, $c(a \cos B - b \cos A) = $

• A. $a^2 - b^2$
• B. $\frac{1}{a^2} - \frac{1}{b^2}$
• C. $a^2 + b^2$
• D. $\frac{1}{a^2} + \frac{1}{b^2}$

Hint:

To solve triangle identities quickly without expanding full formulas, assume a symmetric reference state like an equilateral triangle where $a = b = c = 1$ and $A = B = C = 60^\circ$. Substituting these parameters gives $1(1 \cdot \cos 60^\circ - 1 \cdot \cos 60^\circ) = 1(0.5 - 0.5) = 0$. Testing option (A) yields $1^2 - 1^2 = 0$, confirming it instantly!

Answer 1

The Correct Option is A

Step 1: Choose a clean tool.
We simplify $c(a\cos B - b\cos A)$ using the Law of Cosines for $\cos A$ and $\cos B$.
Step 2: Write the cosines.
$\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}$ and $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$.
Step 3: Form $a\cos B$.
$a\cos B = \dfrac{a^2 + c^2 - b^2}{2c}$.
Step 4: Form $b\cos A$.
$b\cos A = \dfrac{b^2 + c^2 - a^2}{2c}$.
Step 5: Subtract inside the bracket.
$a\cos B - b\cos A = \dfrac{(a^2 + c^2 - b^2) - (b^2 + c^2 - a^2)}{2c} = \dfrac{2a^2 - 2b^2}{2c} = \dfrac{a^2 - b^2}{c}$.
Step 6: Multiply by $c$.
$c\left(\dfrac{a^2 - b^2}{c}\right) = a^2 - b^2$, the same result the projection rule gives.
\[ \boxed{a^2 - b^2} \]