Which of the following relations is correct for `0 < theta lt 90^(@)` ?

Options

A. `sin theta = sin^(2) theta`
B. `sin theta lt sin^(2) theta`
C. `sin theta gt sin^(2) theta`
D. `sin theta = sosec^(2) theta`

Answer

To solve the question regarding the relation that holds for \(0 < \theta < 90^\circ\), we will analyze the given options step by step.

### Step 1: Understanding the Functions
We need to compare the values of \(\sin \theta\) and \(\sin^2 \theta\) for angles \(\theta\) in the range \(0 < \theta < 90^\circ\).

### Step 2: Choosing a Value for \(\theta\)
Let's take a specific value for \(\theta\) to analyze the relationship. A common choice is \(\theta = 45^\circ\).

### Step 3: Calculate \(\sin \theta\) and \(\sin^2 \theta\)
1. Calculate \(\sin 45^\circ\):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.7071 \]

2. Calculate \(\sin^2 45^\circ\):
\[ \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} = 0.5 \]

### Step 4: Compare the Values
Now we compare \(\sin 45^\circ\) and \(\sin^2 45^\circ\):
- \(\sin 45^\circ \approx 0.7071\)
- \(\sin^2 45^\circ = 0.5\)

Since \(0.7071 > 0.5\), we conclude that:
\[ \sin \theta > \sin^2 \theta \quad \text{for } \theta = 45^\circ \]

### Step 5: Generalization
We can generalize this result for all angles \(0 < \theta < 90^\circ\):
- The sine function increases in this interval, while the square of the sine function increases at a slower rate. Thus, \(\sin \theta\) will always be greater than \(\sin^2 \theta\) in this range.

### Conclusion
The correct relation for \(0 < \theta < 90^\circ\) is:
\[ \sin \theta > \sin^2 \theta \]