Simplify the given expression
$\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]+\sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right]$

Hint: We start solving this question by dividing the given expression as two parts and simplify them by taking the constants, $\cos \theta $ and $\sin \theta $ which are outside of matrix, into the matrix and multiplying elements inside the matrix with them. Then we get two expressions for the two parts. Then we add them and obtain a single matrix with the trigonometric functions $\cos \theta $ and $\sin \theta $. Then we use the trigonometric identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$, and simplify the given expression to obtain a simplified version of the given expression.


Complete step by step answer:
Let us consider the given expression, $\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]+\sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right]$.
Let us divide the given expression into two parts $\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]$ and $\sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right]$.
Let us consider $\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]$ as A.
Let us consider $\sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right]$ as B.
Now, let us consider A and simplify A.
$\begin{align}
  & \Rightarrow \cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
\end{align}$
Now, let us consider B and simplify it.
$\begin{align}
  & \Rightarrow \sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   \sin \theta \cos \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right] \\
\end{align}$
As, we have simplified A and B, let us add them.
$\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]+\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   \sin \theta \cos \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & \sin \theta \cos \theta -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta +\sin \theta \cos \theta & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\
\end{matrix} \right] \\
 & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & 0 \\
   0 & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\
\end{matrix} \right]...............\left( 1 \right) \\
\end{align}$
Now let us consider the trigonometric identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$.
Using the above identity, we can write the expression in equation (1) as
$\Rightarrow \left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
So, finally after simplification of the given expression we get that
$\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]+\sin \theta \left[ \begin{matrix}
   \sin \theta & -\cos \theta \\
   \cos \theta & \sin \theta \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$
So, value of given expression is $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$.
Hence answer is $\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$.

Note:
Here while solving this problem, one might make a mistake of multiplying the constant only to the first element of the matrix when the matrix is multiplied with a constant. For example, while simplifying $\cos \theta \left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]$, one might write it as $\left[ \begin{matrix}
   {{\cos }^{2}}\theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]$. But it is wrong. So, we need to remember that when a matrix is multiplied by a constant, every element in the matrix is multiplied by the constant.