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

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(A) Let $t = 	an(	heta/2)$. Then $A = \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix}$. The determinant $|A| = 1 + t^2$. $A^{-1} = \frac{1}{1+t^2} \be

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$\begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$

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(A) Let $t = 	an(	heta/2)$. Then $A = \begin{bmatrix} 1 & -t \ t & 1 \end{bmatrix}$. The determinant $|A| = 1 + t^2$. $A^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & t \ -t & 1 \end{bmatrix}$. Now, $A^{-1}B = \frac{1}{1+t^2} \begin{bmatrix} 1 & t \ -t & 1 \end{bmatrix} \begin{bmatrix} 1 & t \ -t & 1 \end{bmatrix} = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & 2t \ -2t & 1-t^2 \end{bmatrix}$. Using the identities $\cos 	heta = \frac{1-t^2}{1+t^2}$ and $\sin 	heta = \frac{2t}{1+t^2}$, we get $A^{-1}B = \begin{bmatrix} \cos 	heta & \sin 	heta \ -\sin 	heta & \cos 	heta \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & -\tan(\theta/2) \\ \tan(\theta/2) & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ then $A^{-1}B$ is equal to

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