$\mathrm{E}(\theta)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
Now $\mathrm{E}(\alpha)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
and $\mathrm{E}(\beta)=\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right]$
Now $\mathrm{E}(\alpha) \mathrm{E}(\beta)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta\end{array}\right]$
$=\left[\begin{array}{lr}\cos \alpha \cdot \cos \beta & \cos \alpha \sin \beta \\ -\sin \alpha \sin \beta & +\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta & -\sin \alpha \sin \beta \\ -\cos \alpha \sin \beta & +\cos \alpha \cos \beta\end{array}\right]$
$=\left[\begin{array}{cc}\cos (\alpha+\beta) & \sin (\alpha+\beta) \\ -\sin (\alpha+\beta) & \cos (\alpha+\beta)\end{array}\right]$
$=\mathrm{E}(\alpha+\beta)$
$\therefore$ Option $(\mathrm{c})$ is correct.