If `x gt 0, a = x cos theta`, and `b = x sin theta`, then `sqrt(a^(2)+b^(2))=`

Options

A. 1
B. x
C. 2x
D. `x(cos theta + sin theta)`

Explanation

To solve the problem, we need to find the value of \(\sqrt{a^2 + b^2}\) given that \(a = x \cos \theta\) and \(b = x \sin \theta\), where \(x > 0\).

### Step-by-step Solution:

1. **Write down the expressions for \(a\) and \(b\)**:
\[ a = x \cos \theta \]
\[ b = x \sin \theta \]

2. **Square both \(a\) and \(b\)**:
\[ a^2 = (x \cos \theta)^2 = x^2 \cos^2 \theta \]
\[ b^2 = (x \sin \theta)^2 = x^2 \sin^2 \theta \]

3. **Add \(a^2\) and \(b^2\)**:
\[ a^2 + b^2 = x^2 \cos^2 \theta + x^2 \sin^2 \theta \]

4. **Factor out \(x^2\)**:
\[ a^2 + b^2 = x^2 (\cos^2 \theta + \sin^2 \theta) \]

5. **Use the Pythagorean identity**:
We know from trigonometric identities that:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Therefore:
\[ a^2 + b^2 = x^2 \cdot 1 = x^2 \]

6. **Take the square root of both sides**:
\[ \sqrt{a^2 + b^2} = \sqrt{x^2} \]

7. **Since \(x > 0\)**, the square root simplifies to:
\[ \sqrt{x^2} = x \]

### Final Answer:
\[ \sqrt{a^2 + b^2} = x \]