Question

If $y = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$, where $0 \le x < \frac{\pi}{2}$, then $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is

• A. $\frac{1}{4}$
• B. $-\frac{1}{4}$
• C. $-\frac{3}{2}$
• D. $\frac{1}{2}$

Hint:

Whenever an expression of the form $\tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$ or $\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ appears, it always simplifies elegantly to $\frac{\pi}{4} + \frac{x}{2}$ within the first quadrant. Its derivative is always a constant $\frac{1}{2}$, entirely independent of whichever specific value of $x$ is requested!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Simplify the expression $\tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$ and find its derivative.

Step 2: Key Formula or Approach:
Within the first quadrant, use the half-angle identities: $1+\sin x = (\sin\frac{x}{2} + \cos\frac{x}{2})^2$ and $1-\sin x = (\sin\frac{x}{2} - \cos\frac{x}{2})^2$. The square root simplifies to a tangent form, which then reduces elegantly via the arctangent addition formula.

Step 3: Detailed Explanation:
The expression simplifies to $\tan^{-1}\left(\frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right) = \tan^{-1}\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) = \frac{\pi}{4} + \frac{x}{2}$. Differentiating this result gives $\frac{d}{dx}\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{1}{2}$, a constant completely independent of $x$.

Step 4: Final Answer:
The derivative is always $\frac{1}{2}$ regardless of the specific value of $x$.