If A = begin{bmatrix} sec theta & -tan theta | vedclass

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(C) Given $A = \begin{bmatrix} \sec 	heta & -	an 	heta \ -	an 	heta & \sec 	heta \end{bmatrix}$.For a $2 	imes 2$ matrix $A = \begin{bmatrix}

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$\frac{\pi}{3}$

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(C) Given $A = \begin{bmatrix} \sec 	heta & -	an 	heta \ -	an 	heta & \sec 	heta \end{bmatrix}$.For a $2 	imes 2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$, $	ext{adj } A = \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus, $	ext{adj } A = \begin{bmatrix} \sec 	heta & 	an 	heta \ 	an 	heta & \sec 	heta \end{bmatrix}$.Given $A + 	ext{adj } A = 4I$, where $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.$\begin{bmatrix} \sec 	heta & -	an 	heta \ -	an 	heta & \sec 	heta \end{bmatrix} + \begin{bmatrix} \sec 	heta & 	an 	heta \ 	an 	heta & \sec 	heta \end{bmatrix} = \begin{bmatrix} 4 & 0 \ 0 & 4 \end{bmatrix}$.Adding the matrices: $\begin{bmatrix} 2 \sec 	heta & 0 \ 0 & 2 \sec 	heta \end{bmatrix} = \begin{bmatrix} 4 & 0 \ 0 & 4 \end{bmatrix}$.Equating the elements: $2 \sec 	heta = 4 \implies \sec 	heta = 2$.Since $\sec 	heta = 2$, $\cos 	heta = \frac{1}{2}$.For $	heta \in (0, \frac{\pi}{2})$, $	heta = \frac{\pi}{3}$.

If $A = \begin{bmatrix} \sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{bmatrix}$, $\theta \in (0, \frac{\pi}{2})$ such that $A + \text{adj } A = 4I$, then $\theta =$

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