If $\theta -\phi =\dfrac{\pi }{2}$ , then show that $\left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\cos }^{2}}\phi & \cos \phi \sin \phi \\
   \cos \phi \sin \phi & {{\sin }^{2}}\phi \\
\end{matrix} \right]=0$ .

Hint: First, we rewrite the expression $\theta -\phi =\dfrac{\pi }{2}$ as $\phi =\theta -\dfrac{\pi }{2}$ . After that, using the two trigonometric identities $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ , we develop two expressions, which are $\sin \phi =-\cos \theta $ and $\cos \phi =\sin \theta $ . After that, we rewrite the matrix as,
$\Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]$
We now multiply the two matrices to get \[\left[ \begin{matrix}
   {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta & -{{\cos }^{3}}\theta \sin \theta +\sin \theta {{\cos }^{3}}\theta \\
   \cos \theta {{\sin }^{3}}\theta -{{\sin }^{3}}\theta \cos \theta & -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\
\end{matrix} \right]\] . Ruling out the like terms, we can arrive at the conclusion.

Complete step by step answer:
The condition that we are given is,
$\theta -\phi =\dfrac{\pi }{2}$
By taking the $\phi $ to the RHS, and $\dfrac{\pi }{2}$ to the LHS, we can write it as,
$\phi =\theta -\dfrac{\pi }{2}$
Now, we know the trigonometric identities, which are,
$\begin{align}
  & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\
 & \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\
\end{align}$
Using these trigonometric identities, we can show that,
$\begin{align}
  & \Rightarrow \sin \phi =\sin \left( \theta -\dfrac{\pi }{2} \right)=\sin \theta \cos \left( -\dfrac{\pi }{2} \right)-\cos \theta \sin \left( -\dfrac{\pi }{2} \right) \\
 & \Rightarrow \sin \phi =-\cos \theta \\
\end{align}$
And that,
$\begin{align}
  & \Rightarrow \cos \phi =\cos \left( \theta -\dfrac{\pi }{2} \right)=\cos \theta \cos \left( -\dfrac{\pi }{2} \right)-\sin \theta \sin \left( -\dfrac{\pi }{2} \right) \\
 & \Rightarrow \cos \phi =\sin \theta \\
\end{align}$
The matrix that we need to simplify and solve is,
 $\left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\cos }^{2}}\phi & \cos \phi \sin \phi \\
   \cos \phi \sin \phi & {{\sin }^{2}}\phi \\
\end{matrix} \right]$
Using the above expressions that we have derived of $\sin \phi $ and $\cos \phi $ , we can rewrite the matrix as,
$\Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]$
Now, we know the multiplication of two matrices of order $2\times 2$ go as,
$\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]=\left[ \begin{matrix}
   ae+bg & af+bh \\
   ce+dg & cf+dh \\
\end{matrix} \right]$
So, multiplying the above matrices, we get,
\[\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\left( \cos \theta \sin \theta \right)}^{2}} & -{{\cos }^{2}}\theta \sin \theta \cos \theta +\cos \theta \sin \theta {{\cos }^{2}}\theta \\
   \cos \theta \sin \theta {{\sin }^{2}}\theta -{{\sin }^{2}}\theta \sin \theta \cos \theta & -{{\left( \cos \theta \sin \theta \right)}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
\end{align}\]
Simplifying the above matrix elements, we get,
\[\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   {{\cos }^{2}}\theta {{\sin }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta & -{{\cos }^{3}}\theta \sin \theta +\sin \theta {{\cos }^{3}}\theta \\
   \cos \theta {{\sin }^{3}}\theta -{{\sin }^{3}}\theta \cos \theta & -{{\cos }^{2}}\theta {{\sin }^{2}}\theta +{{\sin }^{2}}\theta {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
\end{align}\]
Ruling out the like terms in the element spaces, we get,
\[\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\
   -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right] \\
\end{align}\]
Now, we know that the matrix \[\left[ \begin{matrix}
   0 & 0 \\
   0 & 0 \\
\end{matrix} \right]\] is called a null matrix or a zero matrix, and can also be represented by the digit $0$ as a short form.
Thus, we have proved that the matrix $\left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\cos }^{2}}\phi & \cos \phi \sin \phi \\
   \cos \phi \sin \phi & {{\sin }^{2}}\phi \\
\end{matrix} \right]=0$ if $\theta -\phi =\dfrac{\pi }{2}$ .

Note: We can also solve the problem in a similar way, but now we replace $\theta $ with $2\theta $ . To do this, we rewrite the matrix as
$\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\cos }^{2}}\phi & \cos \phi \sin \phi \\
   \cos \phi \sin \phi & {{\sin }^{2}}\phi \\
\end{matrix} \right] \\
 & =\dfrac{1}{4}\left[ \begin{matrix}
   2{{\cos }^{2}}\theta & 2\cos \theta \sin \theta \\
   2\cos \theta \sin \theta & 2{{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   2{{\cos }^{2}}\phi & 2\cos \phi \sin \phi \\
   2\cos \phi \sin \phi & 2{{\sin }^{2}}\phi \\
\end{matrix} \right] \\
\end{align}$
Now, we know
$\begin{align}
  & 2{{\cos }^{2}}\theta =1+\cos 2\theta \\
 & 2{{\sin }^{2}}\theta =1-\cos 2\theta \\
 & 2\sin \theta \cos \theta =\sin 2\theta \\
\end{align}$
Using these, we get,
$\begin{align}
  & \Rightarrow \left[ \begin{matrix}
   {{\cos }^{2}}\theta & \cos \theta \sin \theta \\
   \cos \theta \sin \theta & {{\sin }^{2}}\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   {{\cos }^{2}}\phi & \cos \phi \sin \phi \\
   \cos \phi \sin \phi & {{\sin }^{2}}\phi \\
\end{matrix} \right] \\
 & =\dfrac{1}{4}\left[ \begin{matrix}
   1+\cos 2\theta & \sin 2\theta \\
   \sin 2\theta & 1-\cos 2\theta \\
\end{matrix} \right]\left[ \begin{matrix}
   1+\cos 2\phi & \sin 2\phi \\
   \sin 2\phi & 1-\cos 2\phi \\
\end{matrix} \right] \\
\end{align}$
Solving this in a similar way and incorporating $\phi =\theta -\dfrac{\pi }{2}$ , we get the same required answer.