If the function f($\theta$) is given as $f\left( \theta \right)=\left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right],f\left( \theta \right).f\left( \phi \right)=$
A. $f\left( \theta +\phi \right)$
B. $f\left( \theta .\phi \right)$
C. $f\left( \theta \right)+f\left( \phi \right)$
D. $f\left( \theta -\phi \right)$

Hint: First of all find $f\left( \phi \right)$ by replacing $\theta $ with $\phi $ in the given matrix. Multiply the matrices $f\left( \theta \right)$ and $f\left( \phi \right)$ by using the general rule of multiplication of matrix given as : $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\times \left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]=\left[ \begin{matrix}
   ae+bg & af+bh \\
   ce+dg & cf+dh \\
\end{matrix} \right]$. Now, use these four trigonometric identities to simplify the expression.
$\begin{align}
  & \left( i \right)\cos A\cos B-\sin A\sin B=\cos \left( A+B \right) \\
 & \left( ii \right)\cos A\cos B+\sin A\sin B=\cos \left( A-B \right) \\
 & \left( iii \right)\sin A\cos B+\cos A\sin B=\sin \left( A+B \right) \\
 & \left( iv \right)\sin A\cos B-\cos A\sin B=\sin \left( A-B \right) \\
\end{align}$

Complete step by step answer:
Here, we have been provided with a matrix :
$f\left( \theta \right)=\left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right]$
We have to find the value of $f\left( \theta \right).f\left( \phi \right)$. To do this first we have to find the matrix $f\left( \phi \right)$.
Now, by replacing $\theta $ with $\phi $ in the matrix $f\left( \theta \right)$, we get,
$f\left( \phi \right)=\left[ \begin{matrix}
   \cos \phi & \sin \phi \\
   -\sin \phi & \cos \phi \\
\end{matrix} \right]$
Now, when we have two matrices, $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ and $\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]$, their multiplication is given as :
$\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\times \left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]=\left[ \begin{matrix}
   ae+bg & af+bh \\
   ce+dg & cf+dh \\
\end{matrix} \right]$
Therefore, by applying the above procedure, we get,
$\begin{align}
  & f\left( \theta \right).f\left( \phi \right)=\left[ \begin{matrix}
   \cos \theta & \sin \theta \\
   -\sin \theta & \cos \theta \\
\end{matrix} \right].\left[ \begin{matrix}
   \cos \phi & \sin \phi \\
   -\sin \phi & \cos \phi \\
\end{matrix} \right] \\
 & \Rightarrow f\left( \theta \right).f\left( \phi \right)=\left[ \begin{matrix}
   \cos \theta \cos \phi +\sin \theta \left( -\sin \phi \right) & \cos \theta \sin \phi +\sin \theta \cos \phi \\
   -\sin \theta \cos \phi +\cos \theta \left( -\sin \phi \right) & -\sin \theta \sin \phi +\cos \theta \cos \phi \\
\end{matrix} \right] \\
 & \Rightarrow f\left( \theta \right).f\left( \phi \right)=\left[ \begin{matrix}
   \cos \theta \cos \phi -\sin \theta \sin \phi & \cos \theta \sin \phi +\sin \theta \cos \phi \\
   -\left( \sin \theta \cos \phi +\cos \theta \sin \phi \right) & \cos \theta \cos \phi -\sin \theta \sin \phi \\
\end{matrix} \right] \\
\end{align}$
Now applying the following trigonometric identities in corresponding elements of matrix, we get,
$\left( i \right)\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)$, in element ${{a}_{11}}$ and ${{a}_{22}}$
$\left( ii \right)\cos A\sin B+\sin A\cos B=\sin \left( A+B \right)$, in element ${{a}_{12}}$ and ${{a}_{21}}$
$\Rightarrow f\left( \theta \right).f\left( \phi \right)=\left[ \begin{matrix}
   \cos \left( \theta +\phi \right) & \sin \left( \theta +\phi \right) \\
   -\sin \left( \theta +\phi \right) & \cos \left( \theta +\phi \right) \\
\end{matrix} \right]$
Clearly, we can see that the matrix obtained in R.H.S can be written as $f\left( \theta +\phi \right)$.
Therefore, $f\left( \theta \right).f\left( \phi \right)=f\left( \theta +\phi \right)$.

So, the correct answer is “Option A”.

Note: One may note that there is an easy method to find the correct option. We can assign some particular values to $\theta $ and $\phi $ like ${{0}^{\circ }},{{30}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$ etc and find the value of $f\left( \theta \right).f\left( \phi \right)$. Now, we will check the options one by one by substituting the same particular value of $\theta $ and $\phi $ in them. But remember that this method can only be applied if the options are provided, otherwise you have to use the general method of multiplication of two matrices as used in the above solution.