Consider `A` and `B` as `2xx2` matrices with determinant equal to 1 then
`tr(AB)-tr(A).tr(B)+tr(AB^(-1))+2` is_______

Explanation

To solve the problem, we need to evaluate the expression \( \text{tr}(AB) - \text{tr}(A) \cdot \text{tr}(B) + \text{tr}(AB^{-1}) + 2 \) given that \( A \) and \( B \) are \( 2 \times 2 \) matrices with \( \det(A) = 1 \) and \( \det(B) = 1 \).

### Step-by-Step Solution:

1. **Understanding the Properties of Trace and Determinant**:
- The determinant of a \( 2 \times 2 \) matrix \( A \) can be expressed as \( \det(A) = ad - bc \) for a matrix of the form \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).
- The trace of a matrix \( A \), denoted \( \text{tr}(A) \), is the sum of its diagonal elements, \( \text{tr}(A) = a + d \).

2. **Using the Given Condition**:
- Since \( \det(A) = 1 \) and \( \det(B) = 1 \), we can use the property of determinants and traces in \( 2 \times 2 \) matrices.

3. **Evaluating \( \text{tr}(AB^{-1}) \)**:
- We know that \( B^{-1} \) exists since \( \det(B) \neq 0 \).
- The trace of the product of matrices has the cyclic property: \( \text{tr}(AB) = \text{tr}(BA) \).
- Therefore, \( \text{tr}(AB^{-1}) = \text{tr}(A \cdot B^{-1}) \).

4. **Setting Up the Expression**:
- We need to evaluate:
\[ \text{tr}(AB) - \text{tr}(A) \cdot \text{tr}(B) + \text{tr}(AB^{-1}) + 2 \]

5. **Using the Identity**:
- From the properties of traces and determinants, we can derive that:
\[ \text{tr}(AB) - \text{tr}(A) \cdot \text{tr}(B) + \text{tr}(AB^{-1}) = 0 \]
- This means that:
\[ \text{tr}(AB) - \text{tr}(A) \cdot \text{tr}(B) + \text{tr}(AB^{-1}) = 0 \]

6. **Final Calculation**:
- Adding 2 to both sides of the equation gives:
\[ 0 + 2 = 2 \]

### Conclusion:
Thus, the value of the expression \( \text{tr}(AB) - \text{tr}(A) \cdot \text{tr}(B) + \text{tr}(AB^{-1}) + 2 \) is \( \boxed{2} \).