trigonometric equation involving two variables Show that the equation $\sec^2\theta = \frac{4xy}{(x+y)^2}$ is only possible when $x =y$.
Can we write it this way : $$\cos^2\theta = \frac{(x+y)^2}{4xy}$$ and $$-1 \leq \cos\theta \leq 1 \Longrightarrow -1 \leq \frac{(x+y)^2}{4xy} \leq 1 $$
or $-4xy \leq (x+y)^2 \leq 4xy $  Please guide... 

$$\sec^2\theta = \frac{4xy}{(x+y)^2}$$
$$\implies \tan^2\theta=\sec^2\theta-1$$
$$=\frac{4xy}{(x+y)^2}-1=\frac{4xy-(x+y)^2}{(x+y)^2}=-\frac{(x-y)^2}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
As $\theta$ is real, $\tan^2\theta\ge0$
$\implies -\left(\frac{x-y}{x+y}\right)^2\ge0$
$\implies \left(\frac{x-y}{x+y}\right)^2\le0$
But the square of any real number is $\ge0,$
as $x,y$ are real $\implies \left(\frac{x-y}{x+y}\right)^2\ge0$
So, $\left(\frac{x-y}{x+y}\right)^2=0\implies x-y=0$ for finite $x,y$
