Which of the following relations is correct for `0ltthetalt90^(@)` ?

Options

A. `sin theta =sin^(2)theta`
B. `sinthetaltsin^(2)theta`
C. `sinthetagtsin^(2)theta`
D. `sintheta=cosec theta`

Answer

To solve the question "Which of the following relations is correct for \(0 < \theta < 90^\circ\)?", we will analyze the relationship between \(\sin \theta\) and \(\sin^2 \theta\) for angles in this range.

### Step-by-step Solution:

1. **Understanding the Range of \(\theta\)**:
- We are given that \(0 < \theta < 90^\circ\). This means \(\theta\) is an acute angle.

2. **Finding the Value of \(\sin \theta\)**:
- The sine function is positive in the range of \(0^\circ\) to \(90^\circ\) and varies from \(0\) to \(1\).
- At \(\theta = 0^\circ\), \(\sin 0 = 0\).
- At \(\theta = 90^\circ\), \(\sin 90 = 1\).
- Therefore, for \(0 < \theta < 90^\circ\), we have \(0 < \sin \theta < 1\).

3. **Calculating \(\sin^2 \theta\)**:
- Since \(\sin \theta\) is between \(0\) and \(1\), squaring \(\sin \theta\) will yield a value that is also between \(0\) and \(1\).
- Specifically, \(0 < \sin^2 \theta < 1\).

4. **Comparing \(\sin \theta\) and \(\sin^2 \theta\)**:
- We need to determine whether \(\sin \theta\) is greater than or less than \(\sin^2 \theta\).
- Since \(\sin \theta\) is a positive value less than \(1\), we can analyze the inequality:
- \(\sin \theta > \sin^2 \theta\) can be rewritten as \(\sin \theta - \sin^2 \theta > 0\).
- Factoring gives us \(\sin \theta (1 - \sin \theta) > 0\).
- Here, \(\sin \theta > 0\) and \(1 - \sin \theta > 0\) for \(0 < \sin \theta < 1\), which means the product is positive.

5. **Conclusion**:
- Therefore, we conclude that \(\sin \theta > \sin^2 \theta\) for \(0 < \theta < 90^\circ\).
- The correct relation is found in option number 3.