If `(pi)/(2)lt alpha lt (3pi)/(4)`, then `sqrt(2tan alpha+(1)/(cos^(2)alpha))` is equal to

Options

A. `-1+tan alpha`
B. `-1- tan alpha`
C. `1+tanalpha`
D. `1-tan alpha`

Answer

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

### Step-by-Step Solution:

1. **Rewrite the Expression**:
Start with the given expression:
\[ \sqrt{2 \tan \alpha + \frac{1}{\cos^2 \alpha}} \]
We know that \( \frac{1}{\cos^2 \alpha} = \sec^2 \alpha \). Therefore, we can rewrite the expression as:
\[ \sqrt{2 \tan \alpha + \sec^2 \alpha} \]

2. **Use the Identity for Secant**:
Recall the trigonometric identity:
\[ \sec^2 \alpha = \tan^2 \alpha + 1 \]
Substitute this into the expression:
\[ \sqrt{2 \tan \alpha + \tan^2 \alpha + 1} \]

3. **Rearrange the Expression**:
The expression inside the square root can be rearranged:
\[ \sqrt{\tan^2 \alpha + 2 \tan \alpha + 1} \]
Notice that this is a perfect square:
\[ \sqrt{(\tan \alpha + 1)^2} \]

4. **Remove the Square Root**:
Taking the square root gives:
\[ |\tan \alpha + 1| \]

5. **Determine the Sign**:
Since \( \frac{\pi}{2} < \alpha < \frac{3\pi}{4} \), we know that:
- \( \tan \alpha < 0 \) (as tangent is negative in the second quadrant).
- Specifically, \( \tan \alpha < -1 \) because at \( \alpha = \frac{3\pi}{4} \), \( \tan \alpha = -1 \).

Therefore, \( \tan \alpha + 1 < 0 \).

6. **Final Expression**:
Since \( \tan \alpha + 1 < 0 \), we can remove the absolute value:
\[ |\tan \alpha + 1| = -(\tan \alpha + 1) = -\tan \alpha - 1 \]

### Conclusion:
Thus, the final result is:
\[ -\tan \alpha - 1 \]