The given trigonometric expression \[{{\sec }^{2}}\theta =\dfrac{4xy}{{{\left( x+y \right)}^{2}}}\] is true if and only if:
1. $x+y\ne 0$
2. $x=y,x\ne 0$
3. $x=y$
4. $x\ne 0,y\ne 0$

Hint: For solving this question you should know about the general concepts of trigonometry and the general mathematical calculations. In this problem we will first compare the value of ${{\sec }^{2}}\theta $ with 1 and then we will use that condition for finding the solution of our original condition.

Complete step-by-step solution:
According to the question it is asked to us whether \[{{\sec }^{2}}\theta =\dfrac{4xy}{{{\left( x+y \right)}^{2}}}\] will be valid or not. As we know that the value of ${{\sec }^{2}}\theta $ is always greater than or equal to one, which means that it can be written that ${{\sec }^{2}}\theta \ge 1$. Now if we solve the given equation, then we will get that,
\[{{\sec }^{2}}\theta =\dfrac{4xy}{{{\left( x+y \right)}^{2}}}\ge 1\]
Now we can also write this as follows,
\[\dfrac{4xy}{{{\left( x+y \right)}^{2}}}\ge 1\]
And this will be further simplified as follows,
\[4xy\ge {{\left( x+y \right)}^{2}}\]
Now if we solve this by taking the right hand side or the RHS to the left hand side or the LHS, thenwe will get as follows,
\[4xy-{{\left( x+y \right)}^{2}}\le 0\]
This can also be written as follows,
\[{{\left( x+y \right)}^{2}}-4xy\le 0\]
Since we know that,
 \[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & \Rightarrow {{x}^{2}}+2xy+{{y}^{2}}-4xy\le 0 \\
 & \Rightarrow {{x}^{2}}-2xy+{{y}^{2}}\le 0 \\
\end{align}\]
Since we also know that,
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
The above expression can be also written as follows,
$\begin{align}
  & {{\left( x-y \right)}^{2}}\le 0 \\
 & \Rightarrow x-y=0 \\
 & \Rightarrow x=y \\
\end{align}$
Now, $x+y=2x$ and also $x+y\ne 0$ (denominator should not be equal to 0), so,
$\begin{align}
  & \Rightarrow 2x\ne 0 \\
 & \Rightarrow x\ne 0 \\
\end{align}$
Hence the correct answer is option 2.

Note: While solving this type of questions, you have to be careful about the values of squares of trigonometric functions. And then solve the given equivalent value of the function very carefully because if it is wrong, then the conditions will also become wrong.