Domain of the function `f (x) = sqrt ((1)/( sin x )-1)` is

Options

A. `uu _(n in I) (2 n pi, 2 n pi + ( pi )/(2))`
B. `uu_(n in I) (2n pi, (2n+1) pi`
C. `uu_(nin I) [(12n -1) pi, 2npi]`
D. None of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{1}{\sin x} - 1} \), we need to ensure that the expression inside the square root is non-negative, as the square root function is only defined for non-negative values.

### Step-by-step Solution:

1. **Set up the inequality**:
We need the expression inside the square root to be greater than or equal to zero:
\[ \frac{1}{\sin x} - 1 \geq 0 \]

2. **Rearrange the inequality**:
Adding 1 to both sides gives:
\[ \frac{1}{\sin x} \geq 1 \]

3. **Multiply both sides by \(\sin x\)**:
Since \(\sin x\) must be positive for this inequality to hold (as multiplying by a negative would reverse the inequality), we multiply both sides by \(\sin x\):
\[ 1 \geq \sin x \]
This is valid only when \(\sin x > 0\).

4. **Determine the range of \(\sin x\)**:
The inequality \(1 \geq \sin x\) is always true since the maximum value of \(\sin x\) is 1. However, we also need \(\sin x\) to be positive:
\[ 0 < \sin x \leq 1 \]

5. **Find the intervals where \(\sin x > 0\)**:
The sine function is positive in the intervals:
\[ (0, \pi) + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
This means that \(\sin x\) is positive in the intervals \( (0, \pi) \), \( (2\pi, 3\pi) \), \( (4\pi, 5\pi) \), etc.

6. **Combine the intervals**:
The domain of the function \( f(x) \) can be expressed as:
\[ x \in (2n\pi, (2n+1)\pi) \quad \text{for } n \in \mathbb{Z} \]

### Final Domain:
Thus, the domain of the function \( f(x) = \sqrt{\frac{1}{\sin x} - 1} \) is:
\[ \bigcup_{n \in \mathbb{Z}} (2n\pi, (2n+1)\pi) \]