The value of `cot^(-1){(sqrt(1-sinx)+sqrt(1+sinx))/(sqrt(1-sinx) -sqrt(1+sinx))} is (0 lt x lt (pi)/(2))`

Options

A. `pi-(x)/(2)`
B. `2pi-x`
C. `(x)/(2)`
D. `2pi-(x)/(2)`

Answer

To solve the problem, we need to find the value of \( \cot^{-1} \left( \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right) \).

### Step-by-Step Solution:

1. **Rewrite the expression**:
We start with the expression:
\[ y = \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \]

2. **Rationalize the denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ y = \frac{(\sqrt{1 - \sin x} + \sqrt{1 + \sin x})^2}{(\sqrt{1 - \sin x})^2 - (\sqrt{1 + \sin x})^2} \]
The denominator simplifies as follows:
\[ (\sqrt{1 - \sin x})^2 - (\sqrt{1 + \sin x})^2 = (1 - \sin x) - (1 + \sin x) = -2\sin x \]

3. **Expand the numerator**:
Now, expand the numerator:
\[ (\sqrt{1 - \sin x} + \sqrt{1 + \sin x})^2 = (1 - \sin x) + (1 + \sin x) + 2\sqrt{(1 - \sin x)(1 + \sin x)} \]
This simplifies to:
\[ 2 + 2\sqrt{1 - \sin^2 x} = 2 + 2\cos x \]

4. **Combine the results**:
So we have:
\[ y = \frac{2 + 2\cos x}{-2\sin x} = -\frac{1 + \cos x}{\sin x} \]

5. **Use cotangent identity**:
Recognizing that:
\[ -\frac{1 + \cos x}{\sin x} = -\cot\left(\frac{x}{2}\right) \]
We can rewrite this as:
\[ y = \cot\left(\frac{x}{2}\right) \text{ (since } \cot^{-1}(-x) = \pi - \cot^{-1}(x) \text{)} \]

6. **Final result**:
Therefore:
\[ \cot^{-1}(y) = \cot^{-1}\left(-\cot\left(\frac{x}{2}\right)\right) = \pi - \frac{x}{2} \]

Thus, the final answer is:
\[ \cot^{-1}\left( \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right) = \pi - \frac{x}{2} \]