Is ` cot^(-1) (-x) = pi - cot^(-1) x , x in R `

Answer

To solve the equation \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \) for \( x \in \mathbb{R} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Left-Hand Side (LHS)**:
We start with the left-hand side of the equation:
\[ \text{LHS} = \cot^{-1}(-x) \]
Let's denote \( \cot^{-1}(-x) \) as \( \theta \). Thus, we have:
\[ \theta = \cot^{-1}(-x) \]
This implies:
\[ \cot(\theta) = -x \]

2. **Analyzing the Quadrants**:
The cotangent function is negative in the second and fourth quadrants. Since the range of \( \cot^{-1}(y) \) is \( (0, \pi) \), \( \theta \) must be in the second quadrant when \( -x \) is positive (i.e., \( x < 0 \)). In this case, we can express \( \theta \) as:
\[ \theta = \pi - \phi \quad \text{where } \phi = \cot^{-1}(x) \]
Therefore, we can write:
\[ \cot(\theta) = \cot(\pi - \phi) = -\cot(\phi) \]

3. **Relating \( \theta \) and \( \phi \)**:
Since \( \cot(\phi) = x \), we can substitute this into our equation:
\[ -\cot(\phi) = -x \]
This confirms that:
\[ \theta = \pi - \cot^{-1}(x) \]

4. **Conclusion**:
Thus, we have shown that:
\[ \cot^{-1}(-x) = \pi - \cot^{-1}(x) \]
for all \( x \in \mathbb{R} \).

### Final Answer:
The statement \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \) is **true** for all \( x \in \mathbb{R} \).

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