If `cos x-sinx=-(5)/(4)`, where `(pi)/(2)ltx lt(3pi)/(4)`, then `cot((x)/(2))` is equal to

Options

A. `(4-sqrt7)/(9)`
B. 8
C. `-8`
D. `(4+sqrt7)/(9)`

Answer

To solve the equation \( \cos x - \sin x = -\frac{5}{4} \) for \( x \) in the interval \( \left( \frac{\pi}{2}, \frac{3\pi}{4} \right) \) and find \( \cot\left(\frac{x}{2}\right) \), we can follow these steps:

### Step 1: Square both sides of the equation
Starting with the equation:
\[ \cos x - \sin x = -\frac{5}{4} \]
Square both sides:
\[ (\cos x - \sin x)^2 = \left(-\frac{5}{4}\right)^2 \]
This simplifies to:
\[ \cos^2 x - 2\cos x \sin x + \sin^2 x = \frac{25}{16} \]

### Step 2: Use the Pythagorean identity
Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ 1 - 2\cos x \sin x = \frac{25}{16} \]
Rearranging gives:
\[ -2\cos x \sin x = \frac{25}{16} - 1 \]
\[ -2\cos x \sin x = \frac{25}{16} - \frac{16}{16} = \frac{9}{16} \]
Thus,
\[ 2\cos x \sin x = -\frac{9}{16} \]

### Step 3: Express \( \cos x + \sin x \)
Now, we need to find \( \cos x + \sin x \). We can use the identity:
\[ (\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2\cos x \sin x \]
Substituting the known values:
\[ (\cos x + \sin x)^2 = 1 + 2\left(-\frac{9}{32}\right) \]
Calculating gives:
\[ (\cos x + \sin x)^2 = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \]
Thus,
\[ \cos x + \sin x = \pm \sqrt{\frac{7}{16}} = \pm \frac{\sqrt{7}}{4} \]

### Step 4: Determine the sign of \( \cos x + \sin x \)
Since \( x \) is in the interval \( \left( \frac{\pi}{2}, \frac{3\pi}{4} \right) \), both \( \cos x \) and \( \sin x \) are positive. Therefore, we take the positive root:
\[ \cos x + \sin x = \frac{\sqrt{7}}{4} \]

### Step 5: Solve for \( \cot\left(\frac{x}{2}\right) \)
We know:
\[ \cot\left(\frac{x}{2}\right) = \frac{1 + \cos x}{\sin x} \]
We have \( \cos x \) from earlier, so we need to find \( \sin x \):
Using the equation \( \cos x - \sin x = -\frac{5}{4} \), we can express \( \sin x \):
\[ \sin x = \cos x + \frac{5}{4} \]
Substituting \( \cos x \):
\[ \sin x = \frac{\sqrt{7}}{4} + \frac{5}{4} = \frac{\sqrt{7} + 5}{4} \]

Now substituting back into the cotangent formula:
\[ \cot\left(\frac{x}{2}\right) = \frac{1 + \cos x}{\sin x} = \frac{1 + \frac{\sqrt{7}}{4}}{\frac{\sqrt{7} + 5}{4}} = \frac{\frac{4 + \sqrt{7}}{4}}{\frac{\sqrt{7} + 5}{4}} = \frac{4 + \sqrt{7}}{\sqrt{7} + 5} \]

### Final Result
Thus, the value of \( \cot\left(\frac{x}{2}\right) \) is:
\[ \cot\left(\frac{x}{2}\right) = \frac{4 + \sqrt{7}}{\sqrt{7} + 5} \]