If `f(x)` is defined on `(0,1)`, then the domain of `f(sinx)` is

Options

A. `(2npi, (2n+ 1)pi), ne in Z`
B. `((2n + 1) (pi)/(2), ( 2n + 3)(pi)/(2)), n in Z`
C. `((n - 1 ) pi, (n + 1) pi), n in Z`
D. none of these

Explanation

To determine the domain of \( f(\sin x) \) given that \( f(x) \) is defined on the interval \( (0, 1) \), we need to find the values of \( x \) for which \( \sin x \) falls within this interval.

### Step-by-Step Solution:

1. **Identify the Range of \( \sin x \)**:
The sine function, \( \sin x \), oscillates between -1 and 1 for all real numbers \( x \). Therefore, the range of \( \sin x \) is:
\[ -1 \leq \sin x \leq 1 \]

2. **Determine When \( \sin x \) is in the Interval \( (0, 1) \)**:
We need to find the values of \( x \) such that:
\[ 0 < \sin x < 1 \]
This means we are looking for the intervals where \( \sin x \) is positive and less than 1.

3. **Find the Intervals for \( \sin x > 0 \)**:
The sine function is positive in the intervals:
\[ (0, \pi) + 2n\pi \quad \text{for any integer } n \]
This means \( x \) can take values in intervals like:
\[ (0, \pi), (2\pi, 3\pi), (4\pi, 5\pi), \ldots \quad \text{and similarly in the negative direction} \]

4. **Find the Intervals for \( \sin x < 1 \)**:
The sine function equals 1 at specific points:
\[ x = \frac{\pi}{2} + 2n\pi \quad \text{for any integer } n \]
Therefore, \( \sin x < 1 \) for all \( x \) except at these points.

5. **Combine the Conditions**:
The combined condition for \( 0 < \sin x < 1 \) gives us the intervals:
\[ (0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi) + 2n\pi \]
This can be expressed as:
\[ (2n\pi, \frac{(2n+1)\pi}{2}) \cup (\frac{(2n+1)\pi}{2}, (2n+1)\pi) \quad \text{for any integer } n \]

6. **Final Domain**:
The domain of \( f(\sin x) \) is thus:
\[ (2n\pi, (2n+1)\pi) \quad \text{for any integer } n \]

### Conclusion:
The domain of \( f(\sin x) \) is given by the intervals derived above, which corresponds to option A in the question.