Question

The value of \[ \sqrt{\sin^4x+4\cos^2x} - \sqrt{\cos^4x+4\sin^2x} \] is

• A. \(1-\cos2x\)
• B. \(\tan2x\)
• C. \(\sin2x\)
• D. \(\cos2x\)

Hint:

Expressions of the form \[ a^2-4a+4 \] can be written as \[ (a-2)^2. \] This simplification is very useful in radical expressions involving trigonometric functions.

Answer 1

The Correct Option is D

Step 1: Simplify the first square root.
Consider \[ \sin^4x+4\cos^2x \] Using \[ \cos^2x=1-\sin^2x, \] we get \[ \sin^4x+4(1-\sin^2x) \] Thus, \[ =\sin^4x-4\sin^2x+4 \] This becomes \[ =(\sin^2x-2)^2 \] Therefore, \[ \sqrt{\sin^4x+4\cos^2x} = \sqrt{(\sin^2x-2)^2} \] Since \[ \sin^2x\leq 1, \] we have \[ \sin^2x-2\lt 0 \] Hence, \[ \sqrt{(\sin^2x-2)^2}=2-\sin^2x \]

Step 2: Simplify the second square root.
Now consider \[ \cos^4x+4\sin^2x \] Using \[ \sin^2x=1-\cos^2x, \] we get \[ \cos^4x+4(1-\cos^2x) \] So, \[ =\cos^4x-4\cos^2x+4 \] Thus, \[ =(\cos^2x-2)^2 \] Hence, \[ \sqrt{\cos^4x+4\sin^2x} = 2-\cos^2x \]

Step 3: Find the difference.
Therefore, \[ (2-\sin^2x)-(2-\cos^2x) \] \[ =\cos^2x-\sin^2x \] Using the identity, \[ \cos2x=\cos^2x-\sin^2x, \] we get \[ \cos2x \]

Step 4: Final conclusion.
Hence, \[ \boxed{\cos2x} \]