Answer:
LHS \[=\text{ }I\text{ }+\text{ }A\] RHS \[=(I\,-A)\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1-0 & 0+\tan \alpha /2 \\ -\tan \alpha /2 & 1-0 \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} 1 & \tan \alpha /2 \\ -\tan \alpha /2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos \alpha +\tan \frac{\alpha }{2}\sin \alpha & -sin\alpha +\tan \frac{\alpha }{2}\cos \alpha \\ -\tan \frac{\alpha }{2}\cos \alpha +\sin \alpha & \tan \frac{\alpha }{2}\sin \alpha +\cos \alpha \\ \end{matrix} \right]\] [multiplying rows by columns] \[\left[ \begin{align} & \because \,\,\,\cos (A-B)=cosA\,cosB+sinA\,sinB \\ & and\,\,sin(A-B)=sinA\,cosB\,-\cos A\,\sin B \\ \end{align} \right]\] = LHS Hence proved.
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