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 Question:
We must differentiate an inverse trigonometric function. Before differentiating, it is crucial to simplify the argument using half-angle trigonometric identities.

Step 2: Detailed Explanation:
Let's simplify the expression under the square root:
We know that $1 = \cos^2(x/2) + \sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$.
Therefore:
$1 + \sin x = (\cos(x/2) + \sin(x/2))^2$
$1 - \sin x = (\cos(x/2) - \sin(x/2))^2$
Substitute these into the function:
$y = \tan^{-1} \left( \sqrt{ \frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))^2} } \right)$
Since $0 \le x < \pi/2$, both $\cos(x/2)$ and $\cos(x/2) - \sin(x/2)$ are positive, so we can drop the square root directly:
$y = \tan^{-1} \left( \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} \right)$
Divide the numerator and denominator by $\cos(x/2)$:
$y = \tan^{-1} \left( \frac{1 + \tan(x/2)}{1 - \tan(x/2)} \right)$
This matches the standard tangent addition formula $\tan(A+B)$:
$y = \tan^{-1} \left( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right)$
Because the argument falls within the principal domain of inverse tangent:
$y = \frac{\pi}{4} + \frac{x}{2}$
Now, differentiate $y$ with respect to $x$:
$y' = \frac{d}{dx} \left( \frac{\pi}{4} + \frac{x}{2} \right)$
$y' = 0 + \frac{1}{2} = \frac{1}{2}$
Since the derivative is a constant $\frac{1}{2}$, its value at $x = \frac{\pi}{6}$ is also $\frac{1}{2}$.

Step 3: Final Answer:
The derivative $y'(\pi/6)$ is $\frac{1}{2}$, matching option (d).