Question

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

• A. $x+y=0$
• B. $x\ne y$
• C. $|x|=|y|\ne 0$
• D. none of these

Answer 1

The correct option is C $|x|=|y|\ne 0$

As $-1\le \sin \theta \le 1$

$\Rightarrow 0\le {\sin }^2\theta \le 1$

$\Rightarrow 0\le \frac{x^2+y^2}{2xy}\le 1$

Now using
$\frac{x^2+y^2}{2xy}\ge 0\Rightarrow x^2+y^2\ge 0$ ...(1)

And $\frac{x^2+y^2}{2xy}\le 1\Rightarrow x^2+y^2\le 2xy$

$\Rightarrow {(x-y)}^2\le 0\Rightarrow x\le y$ ...(2)

From (1) and (2), we get

$x=y$

and $x,y\ne 0$