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

Solution :
$${\left( {x + y} \right)^2} - 4xy = {\left( {x - y} \right)^2} \geqslant 0$$
$$\therefore \,\,{\left( {x + y} \right)^2} \geqslant 4xy$$    and equality holds when $$x = y.$$
But $${\sec ^2}\theta \geqslant 1.$$   Hence, $${\sec ^2}\theta = \frac{{4xy}}{{{{\left( {x + y} \right)}^2}}}$$    can hold only if $${\left( {x + y} \right)^2} = 4xy.$$