To prove ; $\cos \mathrm{\theta }\cdot \left[\begin{array}{cc}\cos \mathrm{\theta }& \sin \mathrm{\theta }\\ -\sin \mathrm{\theta }& \cos \mathrm{\theta }\end{array}\right]+\sin \mathrm{\theta }\left[\begin{array}{cc}\sin \mathrm{\theta }& -\cos \mathrm{\theta }\\ \cos \mathrm{\theta }& \sin \mathrm{\theta }\end{array}\right]=\mathrm{I}$

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

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

Since sin²θ + cos²θ = 1

$=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]=1=\text{ RHS }$

where I = identify matrix

Hence proved.