Checking the equality of an equation(Trigonometery)

Question

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

Answer 1

$$\cos^2\theta=\frac{(x+y)^2}{4xy}\iff \sec^2\theta=\frac{4xy}{(x+y)^2}$$

$$\implies \tan^2\theta=\frac{4xy}{(x+y)^2}-1=-\frac{(x+y)^2-4xy}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$

$$\implies \tan^2\theta+\left(\frac{x-y}{x+y}\right)^2=0$$

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?

Answer 2

Here is a start. We have

$$ 0\leq \cos^2(\theta) \leq 1 \implies 0\leq \frac{(x+y)^2}{4xy} \leq 1 \implies 0\leq (x+y)^2 \leq 4xy \implies \dots.$$

Now, expand, simplify and see what you get.

Note:

$$ -1\leq \cos \theta \leq 1 .$$