${\sin }^2\theta =\frac{(x+y{)}^2}{4xy}$, where $x,y\in R$, gives real $\theta$ if and only if

Solution:

We know that

$\begin{array}{l}0\le {\sin }^2\theta \le 1\\ \Rightarrow 0\le \frac{(x+y{)}^2}{4xy}\le 1\\ \Rightarrow 0\le (x+y{)}^2\le 4xy [\because 4xy>0]\\ \Rightarrow (x-y{)}^2\le 0\Rightarrow x=y\end{array}$

Thus, we have $x=y$ and $xy>0$

$\therefore \left|x\right|=\left|y\right|\ne 0$