If begin{bmatrix} 1 & -tan theta tan theta & | vedclass

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(B) Let $A = \begin{bmatrix} 1 & 	an 	heta \ -	an 	heta & 1 \end{bmatrix}$.Then,$|A| = (1)(1) - (	an 	heta)(-	an 	heta) = 1 + 	an^2 	heta$.

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$a = \cos 2 	heta, b = \sin 2 	heta$

Vedclass · Answer

(B) Let $A = \begin{bmatrix} 1 & 	an 	heta \ -	an 	heta & 1 \end{bmatrix}$.Then,$|A| = (1)(1) - (	an 	heta)(-	an 	heta) = 1 + 	an^2 	heta$.The adjoint of $A$ is $	ext{adj}(A) = \begin{bmatrix} 1 & -	an 	heta \ 	an 	heta & 1 \end{bmatrix}$.We know that $A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{1 + 	an^2 	heta} \begin{bmatrix} 1 & -	an 	heta \ 	an 	heta & 1 \end{bmatrix}$.Given the equation: $\begin{bmatrix} 1 & -	an 	heta \ 	an 	heta & 1 \end{bmatrix} A^{-1} = \begin{bmatrix} a & -b \ b & a \end{bmatrix}$.Substituting $A^{-1}$,we get: $\begin{bmatrix} 1 & -	an 	heta \ 	an 	heta & 1 \end{bmatrix} \frac{1}{1 + 	an^2 	heta} \begin{bmatrix} 1 & -	an 	heta \ 	an 	heta & 1 \end{bmatrix} = \begin{bmatrix} a & -b \ b & a \end{bmatrix}$.Multiplying the matrices: $\frac{1}{1 + 	an^2 	heta} \begin{bmatrix} 1 - 	an^2 	heta & -2 	an 	heta \ 2 	an 	heta & 1 - 	an^2 	heta \end{bmatrix} = \begin{bmatrix} a & -b \ b & a \end{bmatrix}$.This simplifies to: $\begin{bmatrix} \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} & \frac{-2 	an 	heta}{1 + 	an^2 	heta} \ \frac{2 	an 	heta}{1 + 	an^2 	heta} & \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} \end{bmatrix} = \begin{bmatrix} a & -b \ b & a \end{bmatrix}$.Using trigonometric identities $\cos 2 	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$ and $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we get: $\begin{bmatrix} \cos 2 	heta & -\sin 2 	heta \ \sin 2 	heta & \cos 2 	heta \end{bmatrix} = \begin{bmatrix} a & -b \ b & a \end{bmatrix}$.Comparing the elements,we find $a = \cos 2 	heta$ and $b = \sin 2 	heta$.

If $\begin{bmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{bmatrix}^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$,then

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