Question

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

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

Answer 1

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

${\sin }^2\theta =\frac{{(x+y)}^2}{4xy}$
$\because 0\le {\sin }^2\theta \le 1\therefore 0\le \frac{{(x+y)}^2}{4xy}\le 1$
Case 1:
for$x>0;y>0orx<0;y<0$
${(x+y)}^2\ge 0$
and $x<0;y>0orx>0;y<0$
${(x+y)}^2\le 0\Rightarrow x+y=0$
$\therefore x=-y$ ...(1)
Case 2:
${(x+y)}^2-4xy\le 0\Rightarrow {(x-y)}^2\le 0$
$\therefore x=y$ ...(2)
From (1) and (2)
$\left|x\right|=\left|y\right|\ne 0$
Hence, option 'C' is correct.