$\cos \theta \left[ {\begin{array}{*{20}{l}}
  {\cos \theta }&{\sin \theta } \\ 
  { – \sin \theta }&{\cos \theta } 
\end{array}} \right]$ $ + \sin \theta \left[ {\begin{array}{*{20}{c}}
  {\sin \theta }&{ – \cos \theta } \\ 
  {\cos \theta }&{\sin \theta } 
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
  {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\ 
  { – \sin \theta \cos \theta }&{{{\cos }^2}\theta } 
\end{array}} \right]$ $ + \left[ {\begin{array}{*{20}{c}}
  {{{\sin }^2}\theta }&{ – \sin \theta \cos \theta } \\ 
  {\sin \theta \cos \theta }&{{{\sin }^2}\theta } 
\end{array}} \right]$

$=\left[\begin{array}{cc}\cos ^{2} \theta+\sin ^{2} \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta\end{array}\right]$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$      $\left(\because \quad \sin ^{2} \theta=1\right)$