Which of the following pairs of function are identical?

Options

A. `f(x)=e^(In sec^(-1)x) " and " g(x)=sec^(-1)x`
B. `f(x)=tan(tan^(-1)x) " and " g(x)=cot(cot^(-1)x)`
C. f(x)=sgn(x) and g(x)=sgn(sgn(x))
D. `f(x)=cot^(2)*cos^(2)x " and " g(x)=cot^(2)x-cos^(2)x`

Answer

To determine which pairs of functions are identical, we need to analyze each pair of functions given in the question. Two functions \( f(x) \) and \( g(x) \) are identical if \( f(x) = g(x) \) for all \( x \) in their common domain.

### Step-by-Step Solution:

1. **Pair 1: \( f(x) = x \) and \( g(x) = \text{inverse}(x) \)**
- The function \( \text{inverse}(x) \) is typically interpreted as \( \frac{1}{x} \).
- Therefore, \( f(x) = x \) and \( g(x) = \frac{1}{x} \).
- These functions are not identical because \( x \neq \frac{1}{x} \) for all \( x \) (except \( x = 1 \) or \( x = -1 \)).

**Hint:** Check if \( f(x) \) and \( g(x) \) yield the same output for various values of \( x \).

2. **Pair 2: \( f(x) = \tan(\tan^{-1}(x)) \) and \( g(x) = \cot(\cot^{-1}(x)) \)**
- We know that \( \tan(\tan^{-1}(x)) = x \) for all \( x \) in the domain of \( \tan^{-1}(x) \).
- Similarly, \( \cot(\cot^{-1}(x)) = x \) for all \( x \) in the domain of \( \cot^{-1}(x) \).
- Therefore, both functions are identical since \( f(x) = g(x) = x \).

**Hint:** Use the properties of inverse trigonometric functions to simplify.

3. **Pair 3: \( f(x) = \text{signum}(x) \) and \( g(x) = \text{signum}(\text{signum}(x)) \)**
- The signum function, \( \text{signum}(x) \), returns -1 for \( x < 0 \), 0 for \( x = 0 \), and 1 for \( x > 0 \).
- Thus, \( g(x) = \text{signum}(\text{signum}(x)) \) will also return -1, 0, or 1 based on the output of \( \text{signum}(x) \).
- Therefore, both functions are identical since \( f(x) = g(x) \).

**Hint:** Analyze the output of the signum function for different ranges of \( x \).

4. **Pair 4: \( f(x) = \cot^2(x) \cdot \cos^2(x) \) and \( g(x) = \cot^2(x) - \cos^2(x) \)**
- We can rewrite \( f(x) \):
\[ f(x) = \cot^2(x) \cdot \cos^2(x) = \frac{\cos^2(x)}{\sin^2(x)} \cdot \cos^2(x) = \frac{\cos^4(x)}{\sin^2(x)} \]
- For \( g(x) \):
\[ g(x) = \cot^2(x) - \cos^2(x) = \frac{\cos^2(x)}{\sin^2(x)} - \cos^2(x) = \frac{\cos^2(x) - \cos^2(x) \sin^2(x)}{\sin^2(x)} = \frac{\cos^2(x)(1 - \sin^2(x))}{\sin^2(x)} = \frac{\cos^2(x) \cos^2(x)}{\sin^2(x)} = \cot^2(x) \cdot \cos^2(x) \]
- Thus, \( f(x) = g(x) \), making them identical.

**Hint:** Simplify both functions using trigonometric identities.

### Conclusion:
The pairs of functions that are identical are:
- Pair 2: \( f(x) = \tan(\tan^{-1}(x)) \) and \( g(x) = \cot(\cot^{-1}(x)) \)
- Pair 3: \( f(x) = \text{signum}(x) \) and \( g(x) = \text{signum}(\text{signum}(x)) \)
- Pair 4: \( f(x) = \cot^2(x) \cdot \cos^2(x) \) and \( g(x) = \cot^2(x) - \cos^2(x) \)