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

Options

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

Answer

To solve the problem \( \cot^{-1} \left( \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right) \) for \( 0 < x < 2\pi \), we will follow these steps:

### Step 1: Rationalize the Expression
We start by rationalizing the denominator of the expression inside the cotangent inverse. We multiply the numerator and denominator by the conjugate of the denominator:

\[ \text{Let } A = \sqrt{1 - \sin x} \quad \text{and} \quad B = \sqrt{1 + \sin x} \]

So, we have:

\[ \cot^{-1} \left( \frac{A + B}{A - B} \cdot \frac{A + B}{A + B} \right) = \cot^{-1} \left( \frac{(A + B)^2}{A^2 - B^2} \right) \]

### Step 2: Expand the Numerator
Now we expand the numerator:

\[ (A + B)^2 = A^2 + 2AB + B^2 \]

Substituting \( A^2 = 1 - \sin x \) and \( B^2 = 1 + \sin x \):

\[ A^2 + B^2 = (1 - \sin x) + (1 + \sin x) = 2 \]

Now, we find \( 2AB \):

\[ 2AB = 2\sqrt{(1 - \sin x)(1 + \sin x)} = 2\sqrt{1 - \sin^2 x} = 2\cos x \]

Thus, the numerator becomes:

\[ (A + B)^2 = 2 + 2\cos x = 2(1 + \cos x) \]

### Step 3: Simplify the Denominator
Next, we simplify the denominator:

\[ A^2 - B^2 = (1 - \sin x) - (1 + \sin x) = -2\sin x \]

### Step 4: Substitute Back
Now we substitute back into the cotangent inverse:

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

### Step 5: Use Cotangent Identity
Using the identity \( \cot^{-1}(-y) = \pi - \cot^{-1}(y) \):

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

### Step 6: Simplify Further
Now we can express \( \frac{1 + \cos x}{\sin x} \) in terms of tangent:

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

Thus, we have:

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

### Step 7: Final Result
Putting it all together, we get:

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

Thus, the final answer is:

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