Question

Let \(a\) and \(b\) be non-negative real numbers. If \[ \sin x+a\cos x=b, \] then \[ |a\sin x-\cos x|= \] is equal to:

• A. \(\sqrt{a^2-b^2+1}\)
• B. \(\sqrt{b^2-a^2+1}\)
• C. \(\sqrt{1+a^2+b^2}\)
• D. \(\sqrt{a^2+b^2-1}\)

Hint:

Whenever expressions of the form \[ \sin x+a\cos x \] appear, try squaring and combining with another related expression. Also use the identity \[ \sin^2x+\cos^2x=1. \]

Answer 1

The Correct Option is A

Step 1: Square the given equation.
Given \[ \sin x+a\cos x=b \] Squaring both sides, \[ (\sin x+a\cos x)^2=b^2 \] Expanding, \[ \sin^2x+a^2\cos^2x+2a\sin x\cos x=b^2 \]

Step 2: Find \((a\sin x-\cos x)^2\).
Now, \[ (a\sin x-\cos x)^2 \] Expanding, \[ =a^2\sin^2x+\cos^2x-2a\sin x\cos x \] Add the two obtained expressions: \[ (\sin x+a\cos x)^2+(a\sin x-\cos x)^2 \] Substituting, \[ =\sin^2x+a^2\cos^2x+2a\sin x\cos x \] \[ +a^2\sin^2x+\cos^2x-2a\sin x\cos x \] The middle terms cancel. Hence, \[ =(\sin^2x+\cos^2x)+a^2(\sin^2x+\cos^2x) \] Using \[ \sin^2x+\cos^2x=1, \] we get \[ =1+a^2 \] Therefore, \[ b^2+(a\sin x-\cos x)^2=1+a^2 \]

Step 3: Find the required value.
So, \[ (a\sin x-\cos x)^2=1+a^2-b^2 \] Taking square root, \[ |a\sin x-\cos x| = \sqrt{a^2-b^2+1} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{a^2-b^2+1}} \]