Question

If $a \sin\theta + b \cos\theta = c$, then $(a \cos\theta - b \sin\theta)^2 = ?$

• A. $a^2 - b^2 + c^2$
• B. $a^2 + b^2 - c^2$
• C. $a^2 - b^2 - c^2$
• D. $a^2 + b^2 + c^2$

Hint:

The sum of squares of $a\sin\theta+b\cos\theta$ and $a\cos\theta-b\sin\theta$ is always $a^2+b^2$.

Answer 1

The Correct Option is B

Step 1: Concept
Trigonometric identities $(\sin^2\theta + \cos^2\theta = 1)$.

Step 2: Analysis
Let $a \cos\theta - b \sin\theta = x$.
$(a \sin\theta + b \cos\theta)^2 + (a \cos\theta - b \sin\theta)^2 = c^2 + x^2$.

Step 3: Calculation
$(a^2\sin^2\theta + b^2\cos^2\theta + 2ab\sin\theta\cos\theta) + (a^2\cos^2\theta + b^2\sin^2\theta - 2ab\sin\theta\cos\theta) = c^2 + x^2$.
$a^2(1) + b^2(1) = c^2 + x^2 \implies x^2 = a^2 + b^2 - c^2$.

Step 4: Conclusion
The expression equals $a^2 + b^2 - c^2$. Final Answer: (B)