If A = begin{bmatrix} 1 & tan frac{alpha}{2} | vedclass

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(D) Given,$A = \begin{bmatrix} 1 & 	an \frac{\alpha}{2} \ -	an \frac{\alpha}{2} & 1 \end{bmatrix}$ and $AB = I$. Since $AB = I$,we have $B = A^{-1}

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$\cos^2 \frac{\alpha}{2} \cdot A^T$

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(D) Given,$A = \begin{bmatrix} 1 & 	an \frac{\alpha}{2} \ -	an \frac{\alpha}{2} & 1 \end{bmatrix}$ and $AB = I$. Since $AB = I$,we have $B = A^{-1}$. The determinant of $A$ is $|A| = (1)(1) - (	an \frac{\alpha}{2})(-	an \frac{\alpha}{2}) = 1 + 	an^2 \frac{\alpha}{2} = \sec^2 \frac{\alpha}{2}$. The adjoint of $A$ is $	ext{adj}(A) = \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$. Thus,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{\sec^2 \frac{\alpha}{2}} \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$. Since $\frac{1}{\sec^2 \frac{\alpha}{2}} = \cos^2 \frac{\alpha}{2}$,we get $B = \cos^2 \frac{\alpha}{2} \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$. Note that $A^T = \begin{bmatrix} 1 & -	an \frac{\alpha}{2} \ 	an \frac{\alpha}{2} & 1 \end{bmatrix}$. Therefore,$B = \cos^2 \frac{\alpha}{2} \cdot A^T$.

If $A = \begin{bmatrix} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{bmatrix}$ and $AB = I$,then $B$ is equal to

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