If `y=cot^(-1)[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]`, where `0lt xlt(pi)/(2),` then `(dy)/(dx)` is equal to

Options

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

Explanation

To solve the problem, we need to differentiate the function given by:

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

### Step 1: Simplify the Argument of the Cotangent Inverse

We start with the expression inside the cotangent inverse. We can use the identity for sine:

\[ \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]

Thus, we can express \(1 + \sin x\) and \(1 - \sin x\) in terms of sine and cosine of half angles:

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

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

### Step 2: Substitute Back into the Function

Now substituting these back into the expression for \(y\):

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

### Step 3: Recognize the Cotangent Function

This simplifies to:

\[ y = \cot^{-1}\left(\frac{a+b}{a-b}\right) \]

This can be recognized as:

\[ y = \cot\left(\frac{x}{2}\right) \]

### Step 4: Differentiate \(y\)

Now we differentiate \(y\):

\[ \frac{dy}{dx} = \frac{d}{dx} \left(\frac{x}{2}\right) = \frac{1}{2} \]

### Final Answer

Thus, the derivative \(\frac{dy}{dx}\) is:

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