If `(3pi)/4ltalphaltpi," then "sqrt(2cotalpha+1/sin^2alpha)`is equal to

Options

A. `1+cotalpha`
B. `-1-cotalpha`
C. `1-cotalpha`
D. `-1+cotalpha`

Explanation

To solve the problem, we need to evaluate the expression \( \sqrt{2 \cot \alpha + \frac{1}{\sin^2 \alpha}} \) given that \( \frac{3\pi}{4} < \alpha < \pi \).

### Step-by-Step Solution:

1. **Understanding the Range of Alpha**:
Since \( \frac{3\pi}{4} < \alpha < \pi \), we know that in this range, the cotangent function is negative. This is important for our calculations.

2. **Rewriting the Expression**:
We start with the expression:
\[ \sqrt{2 \cot \alpha + \frac{1}{\sin^2 \alpha}} \]
We can rewrite \( \frac{1}{\sin^2 \alpha} \) using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ \frac{1}{\sin^2 \alpha} = \cot^2 \alpha + 1 \]
Therefore, we can substitute this into our expression:
\[ \sqrt{2 \cot \alpha + \cot^2 \alpha + 1} \]

3. **Combining Terms**:
Now we can combine the terms inside the square root:
\[ \sqrt{\cot^2 \alpha + 2 \cot \alpha + 1} \]
This can be factored as:
\[ \sqrt{(\cot \alpha + 1)^2} \]

4. **Taking the Square Root**:
The square root of a square gives us the absolute value:
\[ |\cot \alpha + 1| \]

5. **Considering the Sign**:
Since \( \cot \alpha \) is negative in the range \( \frac{3\pi}{4} < \alpha < \pi \), we have:
\[ \cot \alpha + 1 < 0 \]
Therefore, we can write:
\[ |\cot \alpha + 1| = -(\cot \alpha + 1) = -\cot \alpha - 1 \]

6. **Final Result**:
Thus, the value of the original expression is:
\[ -\cot \alpha - 1 \]

### Conclusion:
The final answer is:
\[ -\cot \alpha - 1 \]