`f(x)=e^(lncot), g(x)=cot^(- 1)x`

Answer

To determine whether the functions \( f(x) = e^{\ln(\cot^{-1}(x))} \) and \( g(x) = \cot^{-1}(x) \) are identical, we will follow these steps:

### Step 1: Understand the Functions
We have:
- \( f(x) = e^{\ln(\cot^{-1}(x))} \)
- \( g(x) = \cot^{-1}(x) \)

### Step 2: Simplify \( f(x) \)
Using the property of logarithms and exponentials, we know that \( e^{\ln(a)} = a \) for \( a > 0 \). Therefore, we can simplify \( f(x) \):
\[ f(x) = e^{\ln(\cot^{-1}(x))} = \cot^{-1}(x) \quad \text{(since \( \cot^{-1}(x) > 0 \))} \]

### Step 3: Compare \( f(x) \) and \( g(x) \)
Now we have:
\[ f(x) = \cot^{-1}(x) \]
\[ g(x) = \cot^{-1}(x) \]
Since both functions are equal, we can conclude that:
\[ f(x) = g(x) \]

### Step 4: Check Domains and Ranges
- **Domain of \( g(x) = \cot^{-1}(x) \)**: The domain is \( (-\infty, \infty) \).
- **Domain of \( f(x) = e^{\ln(\cot^{-1}(x))} \)**: The domain is also \( (-\infty, \infty) \) since \( \cot^{-1}(x) \) is defined for all real \( x \).

- **Range of \( g(x) = \cot^{-1}(x) \)**: The range is \( (0, \pi) \).
- **Range of \( f(x) = \cot^{-1}(x) \)**: The range is also \( (0, \pi) \).

### Conclusion
Since both the domains and ranges are equal, and we have shown that \( f(x) = g(x) \), we conclude that the functions are identical:
\[ \boxed{f(x) = g(x)} \]