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

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(B) Given $A = \begin{bmatrix} 1 & 	an x \ -	an x & 1 \end{bmatrix}$.First,find the transpose $A^{T} = \begin{bmatrix} 1 & -	an x \ 	an x & 1 \e

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

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

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

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