If A = begin{bmatrix} 1 & tan x -tan x & 1 e | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 	an x \ -	an x & 1 \end{bmatrix}$.The determinant $|A| = (1)(1) - (	an x)(-	an x) = 1 + 	an^2 x = \sec^2 x$.T

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$\begin{bmatrix} \cos 2x & -\sin 2x \ \sin 2x & \cos 2x \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & 	an x \ -	an x & 1 \end{bmatrix}$.The determinant $|A| = (1)(1) - (	an x)(-	an x) = 1 + 	an^2 x = \sec^2 x$.The transpose $A^T = \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$.The inverse $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$.Now,$A^T \cdot A^{-1} = \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix} \cdot \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 & -	an x \ 	an x & 1 \end{bmatrix}$$= \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 - 	an^2 x & -	an x - 	an x \ 	an x + 	an x & -	an^2 x + 1 \end{bmatrix}$$= \frac{1}{1 + 	an^2 x} \begin{bmatrix} 1 - 	an^2 x & -2 	an x \ 2 	an x & 1 - 	an^2 x \end{bmatrix}$$= \begin{bmatrix} \frac{1 - 	an^2 x}{1 + 	an^2 x} & \frac{-2 	an x}{1 + 	an^2 x} \ \frac{2 	an x}{1 + 	an^2 x} & \frac{1 - 	an^2 x}{1 + 	an^2 x} \end{bmatrix}$Using trigonometric identities $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$ and $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$,we get:$A^T \cdot A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x \ \sin 2x & \cos 2x \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$,then $A^T \cdot A^{-1} = $

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