If θ is an acute angle such that ${{\sec }^{2}}\theta =3$ , then the value of $\dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta }$ is:
A. $\dfrac{4}{7}$
B. $\dfrac{3}{7}$
C. $\dfrac{2}{7}$
D. $\dfrac{1}{7}$

Hint: The value of expression that we need to find contains ${{\tan }^{2}}\theta $ in the numerator and the denominator so we can write ${{\tan }^{2}}\theta $ as ${{\sec }^{2}}\theta -1$ in that expression. And substitute the value of ${{\sec }^{2}}\theta $ given in the question. Now, we are left with $\cos e{{c}^{2}}\theta $ which can be calculated from ${{\sec }^{2}}\theta $ .

Complete step-by-step answer:
The expression given in the question is:
$\dfrac{{{\tan }^{2}}\theta -\cos e{{c}^{2}}\theta }{{{\tan }^{2}}\theta +\cos e{{c}^{2}}\theta }$
From the trigonometric identities we know that:
$\begin{align}
  & 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\
 & \Rightarrow {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\
\end{align}$
Substituting the value of ${{\tan }^{2}}\theta $ that we have calculated above in the given expression we get,
$\dfrac{{{\sec }^{2}}\theta -1-\cos e{{c}^{2}}\theta }{{{\sec }^{2}}\theta -1+\cos e{{c}^{2}}\theta }$
The value of ${{\sec }^{2}}\theta =3$ is given in the question so plugging this value in the above equation we get,
$\dfrac{3-1-\cos e{{c}^{2}}\theta }{3-1+\cos e{{c}^{2}}\theta }$
$=\dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta }$ ……… Eq. (1)
We can find the value of $\cos e{{c}^{2}}\theta $ from ${{\sec }^{2}}\theta $ as follows:
${{\sec }^{2}}\theta =3$
We know that ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }$ so rewriting the above equation in the following manner.
$\begin{align}
  & \dfrac{1}{{{\cos }^{2}}\theta }=3 \\
 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{1}{3} \\
\end{align}$
From the trigonometric identities we know that,
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
Substituting the value of ${{\cos }^{2}}\theta $ as $\dfrac{1}{3}$ in the above equation we get,
$\begin{align}
  & {{\sin }^{2}}\theta =1-\dfrac{1}{3} \\
 & \Rightarrow {{\sin }^{2}}\theta =\dfrac{2}{3} \\
\end{align}$
From the trigonometric ratios we know that $\cos e{{c}^{2}}\theta $ is the inverse of ${{\sin }^{2}}\theta $ so we can write $\cos e{{c}^{2}}\theta $ as:
$\cos e{{c}^{2}}\theta =\dfrac{3}{2}$
Now, substituting this value of $\cos e{{c}^{2}}\theta $ θ in eq. (1) we get,
$\begin{align}
  & \dfrac{2-\cos e{{c}^{2}}\theta }{2+\cos e{{c}^{2}}\theta } \\
 & =\dfrac{2-\dfrac{3}{2}}{2+\dfrac{3}{2}} \\
 & =\dfrac{1}{7} \\
\end{align}$
From the above calculation, we have calculated the value of the given expression as $\dfrac{1}{7}$ .
Hence, the correct option is (d).

Note: The other way of solving the above problem is as follows:
It is given that ${{\sec }^{2}}\theta =3$ so taking square root on both the sides we get,
$\sec \theta =\sqrt{3}$
Here, we are taking the positive value because θ is given as acute angle.
In the below figure, we have drawn a triangle ABC right angled at B and the figure is also showing an angle $\theta $.