Which of the following pairs of function are identical?

Options

A. `f(x) = sin(tan^(-1)x) , g (x) = x/sqrt(1+x^(2))`
B. `f(x) = sgn ( cot^(-1) x), g(x) = sec^(2) x - tan^(2) x`,where sgn x denotes signum function of x.
C. `f(x) = e ^("In" ( cos^(-1)((x^(2)-1)/(x^(2)+1)))),g(x) = cos^(-1) ((x^(2)-1)/(x^(2)+1))`
D. `f(x)= sin^(-1) ((2x)/(1+x^(2))), g(x) = 2 tan^(-1) x`

Answer

To determine which pairs of functions are identical, we will analyze each pair step by step.

### Step 1: Analyze Option A
Given:
- \( f(x) = \sin(\tan^{-1}(x)) \)
- \( g(x) = \frac{x}{\sqrt{1 + x^2}} \)

**Solution:**
1. Let \( \tan^{-1}(x) = \alpha \). Then, \( x = \tan(\alpha) \).
2. In a right triangle, the opposite side (perpendicular) is \( x \) and the adjacent side (base) is \( 1 \). The hypotenuse can be calculated using the Pythagorean theorem: \( \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \).
3. Therefore, \( \sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1 + x^2}} \).
4. Thus, \( f(x) = \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} = g(x) \).

**Conclusion:** \( f(x) \) and \( g(x) \) are identical.

### Step 2: Analyze Option B
Given:
- \( f(x) = \text{sgn}(\cot^{-1}(x)) \)
- \( g(x) = \sec^2(x) - \tan^2(x) \)

**Solution:**
1. The range of \( \cot^{-1}(x) \) is \( (0, \pi) \), which means \( \cot^{-1}(x) \) is always positive for \( x > 0 \).
2. Therefore, \( \text{sgn}(\cot^{-1}(x)) = 1 \) for \( x > 0 \).
3. We know that \( \sec^2(x) - \tan^2(x) = 1 \) (from the identity \( \sec^2(x) = 1 + \tan^2(x) \)).
4. Thus, \( f(x) = 1 \) and \( g(x) = 1 \).

**Conclusion:** \( f(x) \) and \( g(x) \) are identical.

### Step 3: Analyze Option C
Given:
- \( f(x) = e^{\ln\left(\cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right)\right)} \)
- \( g(x) = \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) \)

**Solution:**
1. By the property of logarithms, \( e^{\ln(a)} = a \). Therefore, \( f(x) = \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) \).
2. Hence, \( f(x) = g(x) \).

**Conclusion:** \( f(x) \) and \( g(x) \) are identical.

### Step 4: Analyze Option D
Given:
- \( f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \)
- \( g(x) = 2\tan^{-1}(x) \)

**Solution:**
1. We know the identity for \( 2\tan^{-1}(x) \):
\[ 2\tan^{-1}(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \]
is true.
2. However, the given \( f(x) \) is \( \sin^{-1}\left(\frac{2x}{1 + x^2}\right) \) and \( g(x) \) is \( 2\tan^{-1}(x) \), which is not equal to \( \sin^{-1}\left(\frac{2x}{1 - x^2}\right) \).
3. Therefore, \( f(x) \neq g(x) \).

**Conclusion:** \( f(x) \) and \( g(x) \) are not identical.

### Final Conclusion
The identical pairs of functions are:
- Option A: \( f(x) = g(x) \)
- Option B: \( f(x) = g(x) \)
- Option C: \( f(x) = g(x) \)