If $\mathrm{A}=\left[\begin{array}{cc}1 & \tan x \\ -\tan x & 1\end{array}\right]$, then $\mathrm{A}^{\mathrm{T}} \mathrm{A}^{-1}=$

The transpose of $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$ is $A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.

The determinant evaluates to $|A| = (1)(1) - (\tan x)(-\tan x) = 1 + \tan^2 x = \sec^2 x$, using the trigonometric identity.

The adjugate matrix is $\begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$, leading to the inverse:

$A^{-1} = \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} = \begin{bmatrix} \cos^2 x & -\sin x \cos x \\ \sin x \cos x & \cos^2 x \end{bmatrix}$

Multiplying matrices:

$A^T A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} \cos^2 x & -\sin x \cos x \\ \sin x \cos x & \cos^2 x \end{bmatrix}$

The product yields $\begin{bmatrix} \cos^2 x - \sin^2 x & -2\sin x \cos x \\ 2\sin x \cos x & \cos^2 x - \sin^2 x \end{bmatrix}$

Applying double-angle identities, this simplifies to $\begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$

This result corresponds to option B.