If \(a\sin\theta+b\cos\theta=c\), prove that \(a\cos\theta-b\sin\theta=\pm\sqrt{a^2+b^2-c^2}\).
\(a\cos\theta-b\sin\theta=\pm\sqrt{a^2+b^2-c^2}\).
Step 1: We are given the relation:
\[ a\sin\theta + b\cos\theta = c. \]
Step 2: Let us consider the expression we want to prove:
\[ a\cos\theta - b\sin\theta. \]
Our goal is to show it equals \( \pm \sqrt{a^2 + b^2 - c^2} \).
Step 3: To connect the two expressions, square and add them:
\[(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2.\]
Step 4: Expand each square.
First part: \((a\sin\theta + b\cos\theta)^2 = a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta.\)
Second part: \((a\cos\theta - b\sin\theta)^2 = a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta.\)
Step 5: Add both results together:
\[ (a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta) \\ + (a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta). \]
Step 6: Simplify the terms.
Step 7: Use the identity \(\sin^2\theta + \cos^2\theta = 1\).
So the sum becomes: \(a^2 + b^2.\)
Step 8: Now replace the first square with the given condition:
Since \(a\sin\theta + b\cos\theta = c\), we have \((a\sin\theta + b\cos\theta)^2 = c^2.\)
Step 9: Therefore,
\[ c^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2. \]
Step 10: Rearranging,
\[ (a\cos\theta - b\sin\theta)^2 = a^2 + b^2 - c^2. \]
Step 11: Taking square roots on both sides,
\[ a\cos\theta - b\sin\theta = \pm \sqrt{a^2 + b^2 - c^2}. \]
Hence Proved.