`f(x)=tan^(-1)x+tan^(-1)(1/x);g(x)=sin^(-1)x+cos^(-1)x` are identical functions if `x in R` (b) `x >0` (c) `x in [-1,1]` (d) `x in [0,1]`

Options

A. `x in R`
B. `x gt 0`
C. `x in [-1, 1]`
D. `x in (0,1]`

Explanation

To determine the conditions under which the functions \( f(x) = \tan^{-1} x + \tan^{-1} \left( \frac{1}{x} \right) \) and \( g(x) = \sin^{-1} x + \cos^{-1} x \) are identical, we will analyze each function step by step.

### Step 1: Analyze \( f(x) \)

1. **Rewrite \( f(x) \)**:
\[ f(x) = \tan^{-1} x + \tan^{-1} \left( \frac{1}{x} \right) \]
We can use the identity \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \) when \( xy < 1 \). Here, we can rewrite \( \tan^{-1} \left( \frac{1}{x} \right) \) as \( \cot^{-1} x \).

2. **Identify the conditions**:
- For \( x > 0 \):
\[ f(x) = \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \]
- For \( x < 0 \):
\[ f(x) = \tan^{-1} x + \cot^{-1} x \text{ is not defined at } x = 0. \]

### Step 2: Analyze \( g(x) \)

1. **Rewrite \( g(x) \)**:
\[ g(x) = \sin^{-1} x + \cos^{-1} x \]
By the identity of inverse trigonometric functions, we know:
\[ g(x) = \frac{\pi}{2} \quad \text{for } x \in [-1, 1]. \]

### Step 3: Compare \( f(x) \) and \( g(x) \)

1. **Identical Functions**:
- For \( f(x) \) to equal \( g(x) \), both functions must equal \( \frac{\pi}{2} \).
- From our analysis, \( f(x) = \frac{\pi}{2} \) when \( x > 0 \), and \( g(x) = \frac{\pi}{2} \) for \( x \in [-1, 1] \).

### Step 4: Determine the range of \( x \)

1. **Combine conditions**:
- \( f(x) \) is defined for \( x > 0 \).
- \( g(x) \) is defined for \( x \in [-1, 1] \).
- The common range where both functions are defined and equal is:
\[ x \in (0, 1]. \]

### Conclusion

The functions \( f(x) \) and \( g(x) \) are identical when \( x \) is in the interval \( [0, 1] \). Therefore, the correct answer is:

**(d) \( x \in [0, 1] \)**.

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