Question

If $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$, then the value of $f'(\frac{\pi}{6})$ is equal to ______.

• A. 0
• B. $\frac{1}{2}$
• C. $-\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Hint:

The factorization $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ is frequently paired with $\sin^2 x + \cos^2 x = 1$ to vastly simplify massive trigonometric fractions!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Simplify the function $f(x)$ by converting all trigonometric ratios to $\sin x$ and $\cos x$.

Step 2: Formula Application:
$\cot x = \frac{\cos x}{\sin x}$ and $\tan x = \frac{\sin x}{\cos x}$.

Step 3: Explanation:
$f(x) = \frac{\sin^2 x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2 x}{1 + \frac{\sin x}{\cos x}} = \frac{\sin^3 x}{\sin x + \cos x} + \frac{\cos^3 x}{\cos x + \sin x}$. $f(x) = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x}$. $f(x) = 1 - \sin x \cos x = 1 - \frac{1}{2}\sin 2x$. $f'(x) = 0 - \frac{1}{2}(\cos 2x \cdot 2) = -\cos 2x$. $f'(\frac{\pi}{6}) = -\cos(2 \cdot \frac{\pi}{6}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$.

Step 4: Final Answer:
$f'(\frac{\pi}{6}) = -\frac{1}{2}$.