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If $\mathrm{A}(\theta)=\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]$ and $\mathrm{AB}=\mathrm{I}$, then $\left(\cos ^{2} \theta\right) \mathrm{B}$ is equal to
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$\mathrm{A}(-\theta)$
$\mathrm{A}(\theta)=\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right] \& \mathrm{AB}=\mathrm{I}$
$\mathrm{B}=\mathrm{I} \mathrm{A}^{-1}, \mathrm{~A}_{11}=1, \mathrm{~A}_{12}=+\tan \theta$
$\mathrm{A}_{21}=-\tan \theta, \mathrm{A}_{22}=1$
$\therefore \mathrm{A}|=| \begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array} \mid=1+\tan ^{2} \theta=\sec ^{2} \theta$
$\Rightarrow \mathrm{A}^{-1}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot \mathrm{A}(-\theta)$
$\Rightarrow\left(\sec ^{2} \theta\right) \cdot \mathrm{B}=\mathrm{A}(-\theta)$
$\Rightarrow\left(\cos ^{2} \theta\right)^{-1} \cdot \mathrm{B}=\mathrm{A}(-\theta)$
$\mathrm{B}=\mathrm{I} \mathrm{A}^{-1}, \mathrm{~A}_{11}=1, \mathrm{~A}_{12}=+\tan \theta$
$\mathrm{A}_{21}=-\tan \theta, \mathrm{A}_{22}=1$
$\therefore \mathrm{A}|=| \begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array} \mid=1+\tan ^{2} \theta=\sec ^{2} \theta$
$\Rightarrow \mathrm{A}^{-1}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$\mathrm{B}=\frac{1}{\sec ^{2} \theta} \cdot \mathrm{A}(-\theta)$
$\Rightarrow\left(\sec ^{2} \theta\right) \cdot \mathrm{B}=\mathrm{A}(-\theta)$
$\Rightarrow\left(\cos ^{2} \theta\right)^{-1} \cdot \mathrm{B}=\mathrm{A}(-\theta)$
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