Question

If $y = \tan^{-1} \left( \sqrt{\frac{1+\sin x}{1-\sin x}} \right)$, $0 \le x < \frac{\pi}{2}$, then $y' \left( \frac{\pi}{6} \right) = $ ______.

• A. $-\frac{1}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Hint:

Never differentiate complex inverse trig functions directly! Always simplify using $\frac{1+\sin x}{1-\sin x} = \tan^2(\frac{\pi}{4} + \frac{x}{2})$ or $\frac{1-\cos x}{1+\cos x} = \tan^2(\frac{x}{2})$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Simplify the expression inside the inverse tangent using trigonometric identities before differentiating.

Step 2: Formula Application:
$1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$
$1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$

Step 3: Explanation:
$\sqrt{\frac{1+\sin x}{1-\sin x}} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} = \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} = \tan(\frac{\pi}{4} + \frac{x}{2})$. Then $y = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$. Differentiating with respect to $x$: $y' = 0 + \frac{1}{2} = \frac{1}{2}$. The value at $x = \frac{\pi}{6}$ remains $\frac{1}{2}$ as it is a constant.

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