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: Understanding the Question:
The problem requires finding the simplified algebraic value of the trigonometric expression $c(a \cos B - b \cos A)$ for a general triangle $ABC$.

Step 2: Key Formula or Approach:
According to the Projection Rule in trigonometry, the sides of any triangle can be expressed in terms of the other sides and their adjacent angles: $$a = b \cos C + c \cos B \implies c \cos B = a - b \cos C$$ $$b = a \cos C + c \cos A \implies c \cos A = b - a \cos C$$ We can distribute the outer variable $c$ and substitute these relations directly.

Step 3: Detailed Explanation:
Let's expand the target expression by distributing the side length $c$: $$E = ac \cos B - bc \cos A$$ Substitute the values from the projection rule formulas: $$E = a(a - b \cos C) - b(b - a \cos C)$$ Distribute the variables $a$ and $b$ through their respective brackets: $$E = a^2 - ab \cos C - b^2 + ab \cos C$$ The complementary cross-terms $-ab \cos C$ and $+ab \cos C$ cancel out completely: $$E = a^2 - b^2$$ This matches the formula given in option (A).

Step 4: Final Answer:
The expression simplifies to $a^2 - b^2$, which corresponds to option (A).