<s>[INST] Checking the equality of an equation(Trigonometery) The equation 
$\displaystyle\cos^2\theta=\frac{(x+y)^2}{4xy}$ is only possible when 
What i need to compare is x and y and derive a relation between them
i.)x

ii.)x=-y
iii.)x>y
iv.)x=y
Pls Answer
 [/INST] $$\cos^2\theta=\frac{(x+y)^2}{4xy}\iff \sec^2\theta=\frac{4xy}{(x+y)^2}$$
$$\implies \tan^2\theta=\frac{4xy}{(x+y)^2}-1=-\frac{(x+y)^2-4xy}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
$$\implies \tan^2\theta+\left(\frac{x-y}{x+y}\right)^2=0$$
As $\theta$ is real,  $\tan^2\theta\ge0$ and $x,y$ are real $\displaystyle\left(\frac{x-y}{x+y}\right)^2\ge0$
Can you take it from here?
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