If A = begin{bmatrix} 1 & cot frac{theta}{2} | vedclass

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(D) Given $A = \begin{bmatrix} 1 & \cot \frac{	heta}{2} \ -\cot \frac{	heta}{2} & 1 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(1) -

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$\left(\frac{1-\cos 	heta}{2}ight) A^T$

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(D) Given $A = \begin{bmatrix} 1 & \cot \frac{	heta}{2} \ -\cot \frac{	heta}{2} & 1 \end{bmatrix}$.First,calculate the determinant $|A| = (1)(1) - (\cot \frac{	heta}{2})(-\cot \frac{	heta}{2}) = 1 + \cot^2 \frac{	heta}{2} = \csc^2 \frac{	heta}{2}$.The adjoint of $A$ is $	ext{adj}(A) = \begin{bmatrix} 1 & -\cot \frac{	heta}{2} \ \cot \frac{	heta}{2} & 1 \end{bmatrix} = A^T$.Thus,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{\csc^2 \frac{	heta}{2}} A^T = \sin^2 \frac{	heta}{2} A^T$.Since $\sin^2 \frac{	heta}{2} = \frac{1-\cos 	heta}{2}$,the correct option is $D$.

If $A = \begin{bmatrix} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{bmatrix}$,then $A^{-1} =$

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