$${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$    is true if and only if

Solution :
We have $${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$
But $${\sec ^2}\theta \geqslant 1$$
$$\eqalign{ & \Rightarrow \,\,\frac{{4xy}}{{{{\left( {x + y} \right)}^2}}} \geqslant 1 \cr & \Rightarrow \,\,4xy \geqslant {x^2} + {y^2} + 2xy \cr & \Rightarrow \,\,{x^2} + {y^2} - 2xy \leqslant 0 \cr & \Rightarrow \,\,{\left( {x - y} \right)^2} \leqslant 0 \cr} $$
⇒ $$x - y = 0$$   [as perfect square of real number can never be negative]
Also then $$x \ne 0$$  as then $${\sec ^2}\theta $$  will become indeterminate.