If A = begin{bmatrix} 1 & tan(theta/2) -tan( | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 	an(	heta/2) \ -	an(	heta/2) & 1 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(1) - (	an(\

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$\cos^2(	heta/2) \cdot A^T$

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(B) Given $A = \begin{bmatrix} 1 & 	an(	heta/2) \ -	an(	heta/2) & 1 \end{bmatrix}$.First,calculate the determinant of $A$:$|A| = (1)(1) - (	an(	heta/2))(-	an(	heta/2)) = 1 + 	an^2(	heta/2) = \sec^2(	heta/2)$.Since $AB = I$,$B$ is the inverse of $A$,i.e.,$B = A^{-1}$.The inverse of a $2 	imes 2$ matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$ is $\frac{1}{|A|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$.Thus,$B = \frac{1}{\sec^2(	heta/2)} \begin{bmatrix} 1 & -	an(	heta/2) \ 	an(	heta/2) & 1 \end{bmatrix}$.Since $\frac{1}{\sec^2(	heta/2)} = \cos^2(	heta/2)$ and the transpose $A^T = \begin{bmatrix} 1 & -	an(	heta/2) \ 	an(	heta/2) & 1 \end{bmatrix}$,we get:$B = \cos^2(	heta/2) \cdot A^T$.

If $A = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ and $AB = I$,then $B = $

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