If $A=\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$ then:

Transpose of $A$ is: $A'=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}$ Now, $A'A=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{bmatrix}$ $A'A=\begin{bmatrix}\cos^2\alpha+\sin^2\alpha & 0 \\ 0 & \sin^2\alpha+\cos^2\alpha\end{bmatrix}$ Since $\sin^2\alpha+\cos^2\alpha=1$ $A'A=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I$