Question

If \( y = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x} \), then \( y'(x) \) is equal to

• A. \( 2\cos^2 x \)
• B. \( 2\cos^3 x \)
• C. \( -\cos 2x \)
• D. \( \cos 2x \)
• E. \( 3\cos x \)

Hint:

Convert expressions into simpler algebraic forms before differentiating.

Answer 1

The Correct Option is D

Concept: Use identities: \[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \]

Step 1: Simplify first term.
\[ \frac{\sin^2 x}{1+\cot x} = \frac{\sin^2 x}{1 + \frac{\cos x}{\sin x}} = \frac{\sin^3 x}{\sin x + \cos x} \]

Step 2: Simplify second term.
\[ \frac{\cos^2 x}{1+\tan x} = \frac{\cos^3 x}{\sin x + \cos x} \]

Step 3: Add both terms.
\[ y = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} \]

Step 4: Use identity: \[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \] \[ y = \sin^2 x - \sin x \cos x + \cos^2 x = 1 - \sin x \cos x \]

Step 5: Differentiate.
\[ y' = -(\cos^2 x - \sin^2 x) = -\cos 2x \]