STATEMENT -1 : The value of `tan^(-1)x+tan^(-1)(1/x)=pi/2, AA x in R -{0}`. and STATEMENT -2 : The value of `tan^(-1).(1/x)={:{(cot^(-1)x,x gt0),(-pi+cot^(-1)x,x lt0):}`

Options

A. Statement -1 is True, Statement-2 is True, Statement -2 is a correct explanation for Statement -4
B. Statement -1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statement -4
C. Statement-1 is True, Statement -2 is False
D. Statement -1 is False, Statement -2 is True

Answer

To solve the given statements step by step, we will analyze both Statement 1 and Statement 2 regarding the properties of inverse trigonometric functions.

### Step-by-Step Solution:

**Statement 1:**
We need to evaluate the expression \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \).

1. **Understanding the Function:**
The function \( \tan^{-1}(x) \) gives the angle whose tangent is \( x \). The function \( \tan^{-1}\left(\frac{1}{x}\right) \) gives the angle whose tangent is \( \frac{1}{x} \).

2. **Using the Identity:**
We know that:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{if } xy < 1 \]
In our case, let \( y = \frac{1}{x} \). Then:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}}\right) = \tan^{-1}\left(\frac{x + \frac{1}{x}}{0}\right) \]
This expression is undefined when \( x = 0 \), but we are considering \( x \in \mathbb{R} - \{0\} \).

3. **Evaluating the Limit:**
As \( x \to 0^+ \) (approaching from the right), \( \tan^{-1}(x) \to 0 \) and \( \tan^{-1}\left(\frac{1}{x}\right) \to \frac{\pi}{2} \). Thus:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \to 0 + \frac{\pi}{2} = \frac{\pi}{2} \]
As \( x \to 0^- \) (approaching from the left), \( \tan^{-1}(x) \to 0 \) and \( \tan^{-1}\left(\frac{1}{x}\right) \to -\frac{\pi}{2} \). Thus:
\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) \to 0 - \frac{\pi}{2} = -\frac{\pi}{2} \]
Therefore, the statement \( \tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \) is only true for \( x > 0 \).

**Conclusion for Statement 1:**
The statement is partially true but not universally true for all \( x \in \mathbb{R} - \{0\} \). Thus, **Statement 1 is false**.

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**Statement 2:**
We need to evaluate \( \tan^{-1}\left(\frac{1}{x}\right) \) based on the sign of \( x \).

1. **Case 1: \( x > 0 \)**
- If \( x > 0 \), then \( \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}(x) \) because \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \).

2. **Case 2: \( x < 0 \)**
- If \( x < 0 \), then \( \tan^{-1}\left(\frac{1}{x}\right) = -\pi + \cot^{-1}(x) \). This is because the angle for \( \tan^{-1}\left(\frac{1}{x}\right) \) will be in the fourth quadrant, which can be represented as \( -\pi + \cot^{-1}(x) \).

**Conclusion for Statement 2:**
The statement correctly describes the relationship of \( \tan^{-1}\left(\frac{1}{x}\right) \) based on the sign of \( x \). Thus, **Statement 2 is true**.

### Final Conclusion:
- Statement 1 is false.
- Statement 2 is true.