Answer
See the full explanation below.
Work Step by Step
Let us consider the left side of the given expression:
$\cos \left( \alpha +\beta \right)\cos \left( \alpha -\beta \right)$
By using the identities of trigonometry,
$\cos \,\left( \alpha +\beta \right)=\cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta $
$\cos \,\left( \alpha -\beta \right)=\cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta $ , the above expression can be further simplified as:
$\begin{align}
& \cos \,\left( \alpha +\beta \right)\cos \,\left( \alpha -\beta \right)=\left( \cos \,\alpha \,\cos \beta -\sin \,\alpha \,\sin \,\beta \right)\times \left( \cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta \right) \\
& ={{\cos }^{2}}\,\alpha \,{{\cos }^{2}}\,\beta +\sin \,\alpha \,\sin \,\beta \cos \,\alpha \,\cos \beta \\
& -\sin \,\alpha \,\sin \,\beta \cos \,\alpha \,\cos \beta -{{\sin }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\
& ={{\cos }^{2}}\,\alpha \,{{\cos }^{2}}\,\beta -{{\sin }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\
& =\left( 1-{{\sin }^{2}}\alpha \right){{\cos }^{2}}\,\beta -{{\sin }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta
\end{align}$
$\begin{align}
& ={{\cos }^{2}}\,\beta -{{\sin }^{2}}\alpha {{\cos }^{2}}\,\beta -{{\sin }^{2}}\,\alpha \,{{\sin }^{2}}\,\beta \\
& ={{\cos }^{2}}\,\beta -{{\sin }^{2}}\alpha \left( {{\cos }^{2}}\,\beta +{{\sin }^{2}}\,\beta \right) \\
& ={{\cos }^{2}}\,\beta -{{\sin }^{2}}\alpha \left( 1 \right) \\
& ={{\cos }^{2}}\,\beta -{{\sin }^{2}}\alpha \\
\end{align}$
Hence, the left side of the given expression is equal to the right side, which is $\cos \left( \alpha +\beta \right)\cos \left( \alpha -\beta \right)={{\cos }^{2}}\,\beta -{{\sin }^{2}}\,\alpha $.
