Trigonometry! 157

We will require that $(x + y) \ne 0$ , for even if $x = y = 0$ the fraction on the RHS will be indeterminate. However, this is not a sufficient condition, since we could have $x = \frac{1}{2}, y = \frac{3}{2}$ yielding $\frac{4xy}{(x + y)^{2}} = \frac{3}{4}$ , which is not possible since we know that $\sec^{2}(\theta) \ge 1$ for all $\theta.$ Now given this last inequality, we see that we must also have

$4xy \ge (x + y)^{2} = x^{2} + y^{2} + 2xy \Longrightarrow 2xy \ge x^{2} + y^{2}.$

Now as $\sec^{2}(\theta) \ge 1$ we see that $x$ and $y$ must have the same sign, so in essence we only need to look at the case where $x,y$ are positive. But in that case, by the AM-GM inequality, we know that $x^{2} + y^{2} \ge 2xy.$ These two inequalities then imply that

$x^{2} + y^{2} = 2xy \Longrightarrow (x - y)^{2} = 0 \Longrightarrow x = y.$

Now we have already noted the requirement that $x + y \ne 0$ , so since we also require that $x = y$ it can be inferred that $x \ne 0.$ Thus the option $\boxed{x = y, x \ne 0}$ is the correct choice.