Which of the following is correct ?

Options

A. `(tan^(-1)x-tan^(-1)y)/(x-y)le1 AAx,y in R,(xney)`
B. `(sin^(-1)x-sin^(-1)y)/(x-y)gt1 AAx, y in [-1,1], x ney`
C. `(cos^(-1)x-cos^(-1)y)/(x-y)lt1 AAx, y in [-1,1], x ney`
D. `(cot^(-1)x-cot^(-1)y)/(x-y)lt1 AAx, y in R, x ney`

Explanation

To solve the problem, we need to analyze the given options related to the inverse trigonometric functions and determine which statements are correct based on the Mean Value Theorem (MVT).

### Step-by-Step Solution:

1. **Understanding the Functions**:
We have the following inverse trigonometric functions:
- \( \tan^{-1}(x) \)
- \( \sin^{-1}(x) \)
- \( \cos^{-1}(x) \)
- \( \cot^{-1}(x) \)

These functions are continuous and differentiable in their respective domains.

2. **Applying the Mean Value Theorem (MVT)**:
The Mean Value Theorem states that if \( f \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \in (a, b) \) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

3. **Analyzing Option A**:
For the function \( f(x) = \tan^{-1}(x) \):
\[ \frac{\tan^{-1}(x) - \tan^{-1}(y)}{x - y} = f'(c) = \frac{1}{1 + c^2} \]
Since \( c \) is between \( x \) and \( y \), \( 1 + c^2 \) is always greater than or equal to 1. Therefore,
\[ \frac{1}{1 + c^2} \leq 1 \]
Thus, option A is correct.

4. **Analyzing Option B**:
For the function \( f(x) = \sin^{-1}(x) \):
\[ \frac{\sin^{-1}(x) - \sin^{-1}(y)}{x - y} = f'(c) = \frac{1}{\sqrt{1 - c^2}} \]
Since \( c \) is between \( x \) and \( y \), \( \sqrt{1 - c^2} \) is always positive and less than or equal to 1. Therefore,
\[ \frac{1}{\sqrt{1 - c^2}} \leq 1 \]
Thus, option B is also correct.

5. **Analyzing Option C**:
For the function \( f(x) = \cos^{-1}(x) \):
\[ \frac{\cos^{-1}(x) - \cos^{-1}(y)}{x - y} = f'(c) = \frac{-1}{\sqrt{1 - c^2}} \]
This derivative is negative, and since \( \sqrt{1 - c^2} \) is always positive, we have:
\[ \frac{-1}{\sqrt{1 - c^2}} \leq 0 \]
However, according to the question, this should be less than 1, which is not satisfied. Thus, option C is incorrect.

6. **Analyzing Option D**:
For the function \( f(x) = \cot^{-1}(x) \):
\[ \frac{\cot^{-1}(x) - \cot^{-1}(y)}{x - y} = f'(c) = \frac{-1}{1 + c^2} \]
This derivative is also negative, and since \( 1 + c^2 \) is always positive, we have:
\[ \frac{-1}{1 + c^2} \leq 0 \]
Again, according to the question, this should be less than 1, which is not satisfied. Thus, option D is also incorrect.

### Conclusion:
The correct options are A and B.