$\text{ If }A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right],\text{ then }A+A^T=\mathrm{I}\text{, if the value of }\alpha \text{ is }$

$$\begin{gathered} \begin{array}{l}\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\\ \Rightarrow {\mathrm{A}}^{\mathrm{\prime }}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\\ \text{ Now, }\mathrm{A}+{\mathrm{A}}^{\mathrm{\prime }}=\mathrm{I}\\ \therefore \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]+\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}2\cos \alpha & 0\\ 0& 2\cos \alpha \end{array}\right]=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right] \\ \begin{array}{l}2\cos \alpha =1\\ \Rightarrow \cos \alpha =\frac{1}{2}=\cos \frac{\pi }{3}\therefore \alpha =\frac{\pi }{3}\end{array}\end{array} \end{gathered}$$