Evaluate without using trigonometric table
$\dfrac{\cos e{{c}^{2}}\left( {{90}^{\circ }}-\theta \right)-{{\tan }^{2}}\left( \theta \right)}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{\cos e{{c}^{2}}{{70}^{\circ }}-{{\tan }^{2}}{{20}^{\circ }}}$

Hint: In this question, we have to evaluate the given trigonometric function. Thus, we will use the trigonometric formulas and identities to get the solution. Thus, first we will apply the trigonometric formula and identity $\cos ec\left( {{90}^{\circ }}-\theta \right)=\sec \theta $ and $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ respectively in the given equation. After the necessary calculations, we will again apply the trigonometric formula $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $ and then apply the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the equation. In the end, we will make the necessary calculations, to get the required solution for the problem.

Complete step by step answer:
According to the problem, we have to evaluate the given trigonometric expression.
Thus, we will use the trigonometric formulas and identities to get the solution.
The expression given to us is $\dfrac{\cos e{{c}^{2}}\left( {{90}^{\circ }}-\theta \right)-{{\tan }^{2}}\left( \theta \right)}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{\cos e{{c}^{2}}{{70}^{\circ }}-{{\tan }^{2}}{{20}^{\circ }}}$ ---- (1)
So, first we will apply the trigonometric formula $\cos ec\left( {{90}^{\circ }}-\theta \right)=\sec \theta $ in equation (1), we get
$= \dfrac{{{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{\cos e{{c}^{2}}\left( {{90}^{\circ }}-{{20}^{\circ }} \right)-{{\tan }^{2}}{{20}^{\circ }}}$
On further solving the above expression, we get
$= \dfrac{{{\sec }^{2}}\left( \theta \right)-{{\tan }^{2}}\left( \theta \right)}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{{{\sec }^{2}}{{20}^{\circ }}-{{\tan }^{2}}{{20}^{\circ }}}$
Now, we will apply the trigonometric identity $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ in the above equation, we get
$= \dfrac{1}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{1}$
Now, we will apply the trigonometric formula $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $in the above equation, we get
 $= \dfrac{1}{4\left( {{\cos }^{2}}\left( {{90}^{\circ }}-{{42}^{\circ }} \right)+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{1}$
On further simplification, we get
$= \dfrac{1}{4\left( {{\sin }^{2}}{{42}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{1}$
Now, we will apply the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above equation, we get
$= \dfrac{1}{4\left( 1 \right)}-\dfrac{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{1}$
Now, we will again apply the trigonometric formula $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ in the above equation, we get
$= \dfrac{1}{4\left( 1 \right)}-\dfrac{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}{{\sin }^{2}}\left( {{90}^{\circ }}-{{52}^{\circ }} \right){{\sec }^{2}}{{52}^{\circ }}}{1}$
$= \dfrac{1}{4\left( 1 \right)}-\dfrac{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}{{\cos }^{2}}{{52}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{1}$
Again, we will apply the trigonometric formula $\sec \theta =\dfrac{1}{\cos \theta }$ in the above equation, we get
$= \dfrac{1}{4\left( 1 \right)}-\dfrac{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}\dfrac{1}{{{\sec }^{2}}{{52}^{\circ }}}{{\sec }^{2}}{{52}^{\circ }}}{1}$
On further solving the above equation, we get
$= \dfrac{1}{4\left( 1 \right)}-\dfrac{2{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}{1}$
$= \dfrac{1}{4}-\dfrac{2}{3}$
Now, we will take the LCM of the denominator in the above equation, we get
$= \dfrac{3-2\left( 4 \right)}{12}$
$= \dfrac{-5}{12}$ which is the required solution
Therefore, for the given equation $\dfrac{\cos e{{c}^{2}}\left( {{90}^{\circ }}-\theta \right)-{{\tan }^{2}}\left( \theta \right)}{4\left( {{\cos }^{2}}{{48}^{\circ }}+{{\cos }^{2}}{{42}^{\circ }} \right)}-\dfrac{2{{\tan }^{2}}{{30}^{\circ }}{{\sin }^{2}}{{38}^{\circ }}{{\sec }^{2}}{{52}^{\circ }}}{\cos e{{c}^{2}}{{70}^{\circ }}-{{\tan }^{2}}{{20}^{\circ }}}$ , its simplified value is equal to $\dfrac{-5}{12}$ .

Note: While solving this problem, do mention all the steps properly to avoid error and confusion. Do mention all the formulas and identity carefully to get an accurate solution. Do not use the trigonometric table to get the solution.