If `tan^(-1) x =y` then :

Options

A. `-1 lt y lt 1`
B. `(-pi)/2 le y le pi/2`
C. `(-pi)/2 lt y lt pi/2`
D. `y in{-(pi)/2,pi/2}`

Answer

To solve the problem, we start with the given equation:

1. **Given**: \( \tan^{-1} x = y \)

2. **Understanding the Function**: The function \( \tan^{-1} x \) (also known as arctan) gives the angle whose tangent is \( x \). The range of the \( \tan^{-1} \) function is important for our solution.

3. **Range of \( \tan^{-1} x \)**: The range of the \( \tan^{-1} x \) function is:
\[ -\frac{\pi}{2} < y < \frac{\pi}{2} \]
This means that the output \( y \) can take values from just above \(-\frac{\pi}{2}\) to just below \(\frac{\pi}{2}\).

4. **Conclusion**: Since \( y = \tan^{-1} x \), we can express the range of \( y \) as:
\[ -\frac{\pi}{2} < y < \frac{\pi}{2} \]

5. **Identifying the Correct Option**: Based on the options provided, we see that:
- Option 1: \( -1 < y < 1 \) (not correct)
- Option 2: \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \) (not correct, as it includes the endpoints)
- Option 3: \( -\frac{\pi}{2} < y < \frac{\pi}{2} \) (correct)
- Option 4: \( y \in (-\frac{\pi}{2}, \frac{\pi}{2}) \) (also correct but not in the form of strict inequalities)

Thus, the correct answer is:
\[ \text{Option 3: } -\frac{\pi}{2} < y < \frac{\pi}{2} \]