If `y= tan ^(-1) sqrt (( 1 - sin x)/( 1 + sin x)) ,` then the vluae of `(dy)/(dx)` at `x = pi /2` is

Options

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

Answer

To find the value of \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\) for the function

\[ y = \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right), \]

we will follow these steps:

### Step 1: Simplify the Argument of the Inverse Tangent Function
We start with the expression inside the inverse tangent:

\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}}. \]

Using the identity for tangent of half angles, we know:

\[ \tan\left(\frac{x}{2}\right) = \frac{\sin x}{1 + \cos x}. \]

Thus, we can express \(1 - \sin x\) and \(1 + \sin x\) in terms of \(\tan\left(\frac{x}{2}\right)\):

\[ 1 - \sin x = \cos^2\left(\frac{x}{2}\right) \quad \text{and} \quad 1 + \sin x = \sin^2\left(\frac{x}{2}\right). \]

This gives us:

\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} = \sqrt{\frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}} = \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = \cot\left(\frac{x}{2}\right). \]

Thus, we can rewrite \(y\) as:

\[ y = \tan^{-1}(\cot\left(\frac{x}{2}\right)). \]

### Step 2: Use the Cotangent Identity
Using the identity \(\tan^{-1}(\cot \theta) = \frac{\pi}{2} - \theta\), we have:

\[ y = \frac{\pi}{2} - \frac{x}{2}. \]

### Step 3: Differentiate \(y\) with Respect to \(x\)
Now we differentiate \(y\):

\[ \frac{dy}{dx} = -\frac{1}{2}. \]

### Step 4: Evaluate at \(x = \frac{\pi}{2}\)
Since \(\frac{dy}{dx} = -\frac{1}{2}\) is a constant, it does not depend on \(x\). Therefore, at \(x = \frac{\pi}{2}\):

\[ \frac{dy}{dx} = -\frac{1}{2}. \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\) is:

\[ \boxed{-\frac{1}{2}}. \]