Evaluate the following expression, ${{\cos }^{2}}30{}^\circ +{{\cos }^{2}}45{}^\circ +{{\cos }^{2}}60{}^\circ +{{\cos }^{2}}90{}^\circ $.

Hint:To solve this question, we should know few values of standard angle like $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\cos 60{}^\circ =\dfrac{1}{2}$ and $\cos 90{}^\circ =0$. By using these values in the given expression, we can get the desired result.

Complete step-by-step answer:
In this question, we are asked to evaluate an expression, that is, ${{\cos }^{2}}30{}^\circ +{{\cos }^{2}}45{}^\circ +{{\cos }^{2}}60{}^\circ +{{\cos }^{2}}90{}^\circ $.
We know that we can represent trigonometric ratios as the ratios of two of the three sides of a right angled triangle. So, applying this, we know that we can write $\cos \theta =\dfrac{Base}{Hypotenuse}$.
Now, to solve the given expression, we will put the values of the cos angles, that are, $\cos 30{}^\circ ,\cos 45{}^\circ ,\cos 60{}^\circ $ and $\cos 90{}^\circ $ respectively.
Now, we know that the values of $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\cos 60{}^\circ =\dfrac{1}{2}$, and $\cos 90{}^\circ =0$. So, we can substitute these values in the given expression, and write the expression as,
${{\cos }^{2}}30{}^\circ +{{\cos }^{2}}45{}^\circ +{{\cos }^{2}}60{}^\circ +{{\cos }^{2}}90{}^\circ ={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( 0 \right)}^{2}}$
Now, we know that, ${{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}=\dfrac{3}{4},{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}=\dfrac{1}{2},{{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4},{{0}^{2}}=0$. So, we will substitute these values in the above expression. So, we get the value of the expression as,
$\dfrac{3}{4}+\dfrac{1}{2}+\dfrac{1}{4}+0$
We will now take the LCM of the above values. So, can write the expression as,
$\dfrac{3+2+1+0}{4}$
And we can further simplify it to get the value of the expression as,
$\dfrac{6}{4}=\dfrac{3}{2}$
Hence, we can say that the value of the given expression, ${{\cos }^{2}}30{}^\circ +{{\cos }^{2}}45{}^\circ +{{\cos }^{2}}60{}^\circ +{{\cos }^{2}}90{}^\circ $ is $\dfrac{5}{2}$.

Note: We can think of solving this question by using the identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$, which will give ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $, which is also a correct method of solving, but by using this method, the number of terms will increase, which may increase the possibility of calculation mistakes. So, it is always better to remember the values of $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2},\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\cos 60{}^\circ =\dfrac{1}{2}$, and $\cos 90{}^\circ =0$.