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:
The problem presents an implicit trigonometric function mapping $x$ to $y$. We need to simplify this expression using core trigonometric identities first, and then evaluate its first derivative at the specific point $x = \frac{\pi}{6}$.

Step 2: Key Formula or Approach:
1. Use the half-angle conversions to transform the square root terms: $$1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$$ $$1 - \sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$$ 2. Simplify the inner fraction by dividing both the numerator and denominator by $\cos\frac{x}{2}$ to bring it to a tangent compound angle format: $\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta}$.

Step 3: Detailed Explanation:
Let's substitute the half-angle identity forms inside the radical: $$\sqrt{\frac{1 + \sin x}{1 - \sin x}} = \sqrt{\frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}} = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$$ Note that the signs are positive since $x \in [0, \frac{\pi}{2})$, meaning $\cos\frac{x}{2} > \sin\frac{x}{2}$. Now, divide numerator and denominator by $\cos\frac{x}{2}$: $$\frac{\frac{\cos(x/2)}{\cos(x/2)} + \frac{\sin(x/2)}{\cos(x/2)}}{\frac{\cos(x/2)}{\cos(x/2)} - \frac{\sin(x/2)}{\cos(x/2)}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}$$ Recognize this as the expansion for $\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$. Now substitute this back into our primary equation: $$y = \tan^{-1}\left[\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right]$$ Since $x \in [0, \frac{\pi}{2})$, the angle $\left(\frac{\pi}{4} + \frac{x}{2}\right)$ falls perfectly inside the principal inverse domain $\left(0, \frac{\pi}{2}\right)$, allowing clean cancellation: $$y = \frac{\pi}{4} + \frac{x}{2}$$ Differentiating this linear function with respect to $x$: $$\frac{dy}{dx} = 0 + \frac{1}{2} = \frac{1}{2}$$ Since the derivative is a constant value ($\frac{1}{2}$), its evaluation at $x = \frac{\pi}{6}$ remains exactly $\frac{1}{2}$.

Step 4: Final Answer:
The derivative evaluates to $\frac{1}{2}$, matching option (D).