If `sintheta-costhetalt0, then theta` lies between

Options

A. `n pi-(3pi)/(4)and n pi+(pi)/(4), n inZ`
B. `n pi-(pi)/(4)and n pi+(3pi)/(4), n inZ`
C. `2npi-(3pi)/(4)and 2npi-(pi)/(4),n inZ`
D. `2npi-(3pi)/(4)and 2npi+(pi)/(4),n in Z`

Answer

To solve the problem, we need to determine the range of \(\theta\) given that \(\sin \theta - \cos \theta < 0\).

### Step-by-Step Solution:

1. **Start with the given inequality**:
\[ \sin \theta - \cos \theta < 0 \]

2. **Rearrange the inequality**:
\[ \sin \theta < \cos \theta \]

3. **Divide both sides by \(\sqrt{2}\)** (to help manipulate the expression):
\[ \frac{\sin \theta}{\sqrt{2}} < \frac{\cos \theta}{\sqrt{2}} \]

4. **Recognize that \(\frac{1}{\sqrt{2}} = \sin \frac{\pi}{4} = \cos \frac{\pi}{4}\)**:
\[ \sin \theta \cdot \sin \frac{\pi}{4} < \cos \theta \cdot \cos \frac{\pi}{4} \]

5. **Use the sine subtraction formula**:
\[ \sin \theta \cos \frac{\pi}{4} - \cos \theta \sin \frac{\pi}{4} < 0 \]
This simplifies to:
\[ \sin\left(\theta - \frac{\pi}{4}\right) < 0 \]

6. **Determine where the sine function is negative**:
The sine function is negative in the intervals:
\[ \pi < \theta - \frac{\pi}{4} < 2\pi \]

7. **Add \(\frac{\pi}{4}\) to all parts of the inequality**:
\[ \pi + \frac{\pi}{4} < \theta < 2\pi + \frac{\pi}{4} \]

8. **Convert the bounds to a common fraction**:
\[ \frac{5\pi}{4} < \theta < \frac{9\pi}{4} \]

9. **Express the solution in terms of \(n\)**:
Since the sine function is periodic, we can express the solution as:
\[ 2n\pi - \frac{3\pi}{4} < \theta < 2n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Final Answer:
Thus, the range of \(\theta\) is:
\[ \theta \in \left(2n\pi - \frac{3\pi}{4}, 2n\pi + \frac{\pi}{4}\right), \quad n \in \mathbb{Z} \]