If ` y = tan^(-1) sqrt((1-sinx)/(1+sinx))`, then the value of `(dy)/(dx) "at x = (pi)/(6) ` is

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. 1
D. -1

Explanation

To solve the problem, we need to find the derivative of the function \( y = \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) \) and evaluate it at \( x = \frac{\pi}{6} \).

### Step-by-Step Solution:

1. **Write the function**:
\[ y = \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) \]

2. **Use the identity for \( 1 - \sin x \) and \( 1 + \sin x \)**:
We know that:
\[ 1 - \sin x = 2 \sin^2\left(\frac{x}{2}\right) \quad \text{and} \quad 1 + \sin x = 2 \cos^2\left(\frac{x}{2}\right) \]
Therefore,
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}} = \tan\left(\frac{x}{2}\right) \]

3. **Substitute back into the equation for \( y \)**:
\[ y = \tan^{-1} \left( \tan\left(\frac{x}{2}\right) \right) \]
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range,
\[ y = \frac{x}{2} \]

4. **Differentiate \( y \) with respect to \( x \)**:
\[ \frac{dy}{dx} = \frac{1}{2} \]

5. **Evaluate \( \frac{dy}{dx} \) at \( x = \frac{\pi}{6} \)**:
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Answer:
\[ \frac{dy}{dx} \text{ at } x = \frac{\pi}{6} \text{ is } \frac{1}{2}. \]