Question

${\sin }^2\theta =\frac{4xy}{(x+y{)}^2}$ is true if and only if :

• A. $x+y\ne 0$
• B. $x=y,x\ne 0,y\ne 0$
• C. $x=y$
• D. $x\ne 0,y\ne 0$

Answer 1

Answer: (A)

$\because \sin \theta \le 1$
${\sin }^2\theta \le 1$
$\Rightarrow \frac{4xy}{(x+y{)}^2}\le 1$
$\Rightarrow 0\le (x+y{)}^2-4xy$
$\Rightarrow x^2+y^2+2xy-4xy\ge 0$
$\Rightarrow (x-y{)}^2\ge 0$
Which is true for all real $x$ and $y$ provided $x+y\ne 0,$ otherwise $\frac{4xy}{(x+y{)}^2}$ will be meaningless.
Note $:{\sec }^2\theta \le \frac{4xy}{(x+y{)}^2}$ is possible only, if $x=y$.