If \(A=\left[\begin{array}{cc}1 & \tan \alpha / 2 \\ -\tan \alpha / 2 & 1\end{array}\right]\) and \(A B=I\), then \(B\) is equal to

G̣iven,
\(A=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]\) and \(A B=I\)
\(\Rightarrow B=A^{-1}\)
\(\begin{aligned} & \text { Now } A=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right] \\ & |A|=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]=1+\operatorname{an}^2 \frac{\alpha}{2}=\sec ^2 \frac{\alpha}{2} \\ & \operatorname{adj} A=\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right] \\ & A^{-1}=\frac{1}{\sec ^2 \frac{\alpha}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right] \\ & =\cos ^2 \frac{\alpha}{2}\left[\begin{array}{cc}1 & \left.-\tan \frac{\alpha}{2}\right] \\ \tan \frac{\alpha}{2} & 1\end{array} \cos ^2 \frac{\alpha}{2} A T\right. \\ & \end{aligned}\)
\(B=\left[\cos ^2 \frac{\alpha}{2}\right] A T\)