f(x) and and g(x) are identical or not ? `f(x)=Sec^(-1)x+Cosec^(-1)x,g(x)=pi/2`

Answer

To determine whether the functions \( f(x) \) and \( g(x) \) are identical, we need to analyze the given functions:

1. **Identify the functions**:
- \( f(x) = \sec^{-1}(x) + \csc^{-1}(x) \)
- \( g(x) = \frac{\pi}{2} \)

2. **Evaluate \( f(x) \)**:
- The functions \( \sec^{-1}(x) \) and \( \csc^{-1}(x) \) are defined for \( x \) in certain intervals. Specifically:
- \( \sec^{-1}(x) \) is defined for \( |x| \geq 1 \)
- \( \csc^{-1}(x) \) is defined for \( |x| \geq 1 \)

3. **Find the domain of \( f(x) \)**:
- Since both inverse functions are defined for \( |x| \geq 1 \), the domain of \( f(x) \) is \( (-\infty, -1] \cup [1, \infty) \).

4. **Evaluate \( f(x) \) for specific values**:
- Let's evaluate \( f(x) \) at the boundaries of its domain:
- For \( x = 1 \):
\[ f(1) = \sec^{-1}(1) + \csc^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \]
- For \( x = -1 \):
\[ f(-1) = \sec^{-1}(-1) + \csc^{-1}(-1) = \pi + (-\frac{\pi}{2}) = \frac{\pi}{2} \]

5. **Check if \( f(x) = g(x) \)**:
- Since \( g(x) = \frac{\pi}{2} \) for all \( x \) in its domain, and we found that \( f(1) = \frac{\pi}{2} \) and \( f(-1) = \frac{\pi}{2} \), we need to check if \( f(x) \) is always equal to \( g(x) \) for all \( x \) in the domain of \( f(x) \).

6. **Analyze the behavior of \( f(x) \)**:
- To determine if \( f(x) \) is always equal to \( \frac{\pi}{2} \), we can analyze the general behavior of \( f(x) \):
- As \( x \) increases beyond 1, \( \sec^{-1}(x) \) increases and \( \csc^{-1}(x) \) also increases, leading to \( f(x) > \frac{\pi}{2} \).
- Similarly, as \( x \) decreases below -1, both \( \sec^{-1}(x) \) and \( \csc^{-1}(x) \) will also lead to \( f(x) > \frac{\pi}{2} \).

7. **Conclusion**:
- Since \( f(x) \) is not equal to \( g(x) \) for all \( x \) in the domain of \( f(x) \), we conclude that \( f(x) \) and \( g(x) \) are not identical.

### Final Answer:
The functions \( f(x) \) and \( g(x) \) are not identical.