Evaluate the following expression, $\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ $.

Hint:When we look at the question, we know that the angles are standard angles. So, we should know their values like, $\cos 60{}^\circ =\dfrac{1}{2},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$. By using these values, we can get the answer.

Complete step-by-step answer:
In this question, we have been asked to evaluate the given expression, $\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ $. To solve this question, we will first look at the angles. We can see that the angles are standard angles. For the same we should know the values of the standard angles. Now, we will put those values of sine and cosine angles, which are, $\cos 60{}^\circ =\dfrac{1}{2},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$ in the given expression of the question to get the desired answer. And, therefore, we can write the given expression, $\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ $ as,
$=\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}$
We will simplify it further. So, we get the expression as,
$\dfrac{1}{2\sqrt{2}}-\dfrac{\sqrt{3}}{2\sqrt{2}}$
Now, we will take the LCM. By doing so, we get the value as,
$\dfrac{1-\sqrt{3}}{2\sqrt{2}}$
We can rationalize the denominator by multiplying the numerator and the denominator with $\sqrt{2}$. So, doing that, we get the value as,
$\begin{align}
  & \dfrac{1-\sqrt{3}}{2\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}} \\
 & \Rightarrow \dfrac{\sqrt{2}\left[ 1-\sqrt{3} \right]}{4} \\
\end{align}$
Hence, we can say that the value of the expression given in the question, that is, $\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ $ is $\dfrac{\sqrt{2}\left[ 1-\sqrt{3} \right]}{4}$.

Note: While solving this question, one can think of applying the identity, $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$. If we apply this identity, then $a=60{}^\circ ,b=45{}^\circ $ and we would get (a + b) as, $60{}^\circ +45{}^\circ =105{}^\circ $. And so, according to this identity, we would get, $\cos 105{}^\circ =\cos 60{}^\circ \cos 45{}^\circ -\sin 60{}^\circ \sin 45{}^\circ $ and then we would require the value of $\cos 105{}^\circ $, which would make the question more tougher and lengthier. The possible mistake that one can make is by mixing up the values of $\sin 60{}^\circ $ and $\cos 60{}^\circ $, we could end up taking $\cos 60{}^\circ =\dfrac{\sqrt{3}}{2}$ instead of $\dfrac{1}{2}$. So, it is very important to memorise the values of sine, cosine and tangent at standard angles for solving this type of questions easily in the exams.