`A=[1t a n x-t a n x1]a n df(x)` is defined as `f(x)=d e tdot(A^T A^(-1))` en the value of `(f(f(f(ff(x))))_` is `(ngeq2)` _________.

Explanation

To solve the given problem step by step, we will follow the reasoning presented in the video transcript.

### Step-by-Step Solution:

1. **Define the Matrix A**:
\[ A = \begin{bmatrix} 1 & t \\ a & n \\ x - t & a \\ n & x & 1 \end{bmatrix} \]

2. **Calculate the Determinant of A**:
The determinant of matrix \( A \) can be calculated using the formula for determinants. For the specific matrix given, we can simplify it as follows:
\[ \text{det}(A) = 1 - (-t)(-t) = 1 - t^2 \]
However, based on the video, it seems we have:
\[ \text{det}(A) = 1 + \tan^2(x) \]
We know from trigonometry that:
\[ 1 + \tan^2(x) = \sec^2(x) \]
Therefore:
\[ \text{det}(A) = \sec^2(x) \]

3. **Use Properties of Determinants**:
The determinant of the transpose of a matrix is equal to the determinant of the matrix itself:
\[ \text{det}(A^T) = \text{det}(A) = \sec^2(x) \]

4. **Define the Function f(x)**:
The function \( f(x) \) is defined as:
\[ f(x) = \text{det}(A^T A^{-1}) \]
Using properties of determinants:
\[ f(x) = \text{det}(A^T) \cdot \text{det}(A^{-1}) = \text{det}(A) \cdot \frac{1}{\text{det}(A)} = 1 \]

5. **Evaluate f(f(f(f(x))))**:
Since \( f(x) = 1 \) for any \( x \), we can substitute:
\[ f(f(x)) = f(1) = 1 \]
Continuing this process:
\[ f(f(f(x))) = f(1) = 1 \]
\[ f(f(f(f(x)))) = f(1) = 1 \]

6. **Final Result**:
Therefore, the value of \( f(f(f(f(x)))) \) is:
\[ \boxed{1} \]