Find the domain of: `y=cot^(-1)(sqrt(x^(2)-1))`

Explanation

To find the domain of the function \( y = \cot^{-1}(\sqrt{x^2 - 1}) \), we need to follow these steps:

### Step 1: Identify the conditions for the square root
The expression inside the square root, \( \sqrt{x^2 - 1} \), must be defined. This means that the expression \( x^2 - 1 \) must be greater than or equal to zero:
\[ x^2 - 1 \geq 0 \]

### Step 2: Solve the inequality
We can solve the inequality \( x^2 - 1 \geq 0 \) by factoring:
\[ (x - 1)(x + 1) \geq 0 \]
Next, we find the critical points by setting the factors equal to zero:
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 3: Test intervals around the critical points
We need to test the intervals defined by the critical points \( -1 \) and \( 1 \):
1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ (-2 - 1)(-2 + 1) = (-3)(-1) = 3 \quad (\text{positive}) \]
2. **Interval \( (-1, 1) \)**: Choose \( x = 0 \):
\[ (0 - 1)(0 + 1) = (-1)(1) = -1 \quad (\text{negative}) \]
3. **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[ (2 - 1)(2 + 1) = (1)(3) = 3 \quad (\text{positive}) \]

### Step 4: Determine the intervals where the inequality holds
From the tests, we find that the inequality \( (x - 1)(x + 1) \geq 0 \) holds true in the intervals:
- \( (-\infty, -1] \)
- \( [1, \infty) \)

### Step 5: Combine the intervals
Thus, the domain of \( \sqrt{x^2 - 1} \) is:
\[ (-\infty, -1] \cup [1, \infty) \]

### Step 6: Consider the domain of \( \cot^{-1}(x) \)
The function \( \cot^{-1}(x) \) is defined for all real numbers, so there are no additional restrictions from this function.

### Final Domain
Therefore, the domain of \( y = \cot^{-1}(\sqrt{x^2 - 1}) \) is:
\[ \text{Domain} = (-\infty, -1] \cup [1, \infty) \]