if  $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] then A^2=I$ is true  for 

we are given $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]also A^2=I$

$$\begin{gathered} A^2=A.A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ {A}^2=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & \cos \theta \sin \theta +\sin \theta \cos \theta \\ -\sin \theta \cos \theta -\cos \theta \sin \theta & -{\sin }^2\theta +{\cos }^2\theta \end{array}\right] \\ now A^2=I \\ \Rightarrow {\cos }^2\theta -{\sin }^2\theta =1 \\ \Rightarrow \cos 2\theta =1 \\ \Rightarrow 2\theta =0 \\ \Rightarrow \theta =0 \\ also 2\sin \theta \cos \theta =0 \\ \sin 2\theta =0 \\ \Rightarrow \theta =0 \\ also \\ -2\sin \theta \cos \theta =0 \\ -2\sin 2\theta =0 \\ also \theta =0 \end{gathered}$$

hence for $\theta =0$

$A^2=I$