If `0lt theta lt 90^(@)`, the `(sin theta+cos theta)` is :

Options

A. less than 1
B. equal to 1
C. greater than 1
D. none of the above

Answer

To solve the problem, we need to analyze the expression \( \sin \theta + \cos \theta \) for \( 0 < \theta < 90^\circ \).

### Step-by-Step Solution:

1. **Understanding the Quadrant**:
Since \( \theta \) is between \( 0^\circ \) and \( 90^\circ \), it lies in the first quadrant. In this quadrant, both sine and cosine functions are positive.

**Hint**: Remember that in the first quadrant, both sine and cosine values are positive.

2. **Using the Pythagorean Identity**:
We know from trigonometry that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

3. **Expressing \( \sin \theta + \cos \theta \)**:
We can square the expression \( \sin \theta + \cos \theta \):
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta \]
Substituting the Pythagorean identity:
\[ (\sin \theta + \cos \theta)^2 = 1 + 2\sin \theta \cos \theta \]

4. **Finding the Range of \( 2\sin \theta \cos \theta \)**:
The term \( 2\sin \theta \cos \theta \) can be rewritten using the double angle identity:
\[ 2\sin \theta \cos \theta = \sin(2\theta) \]
Since \( 0 < \theta < 90^\circ \), it follows that \( 0 < 2\theta < 180^\circ \), and thus \( \sin(2\theta) \) is positive and ranges from \( 0 \) to \( 1 \).

5. **Analyzing \( \sin \theta + \cos \theta \)**:
Therefore, we can conclude:
\[ (\sin \theta + \cos \theta)^2 = 1 + \sin(2\theta) \]
Since \( \sin(2\theta) > 0 \) in the first quadrant, we have:
\[ (\sin \theta + \cos \theta)^2 > 1 \]
Taking the square root gives:
\[ \sin \theta + \cos \theta > 1 \]

6. **Conclusion**:
Thus, for \( 0 < \theta < 90^\circ \), we conclude that:
\[ \sin \theta + \cos \theta > 1 \]

### Final Answer:
The correct option is that \( \sin \theta + \cos \theta \) is greater than 1.

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