Question

${\sec }^2\theta$ =$\frac{4xy}{(x+y{)}^2}$, where $x\in R,$ $y\in R$, 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), (B)

The correct options are
A $x+y\ne 0$
B $x=y,x\ne 0,y\ne 0$

Given, ${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}$
${\cos }^2\theta =\frac{{(x+y)}^2}{4xy}$ Where $x\ne 0,y\ne 0$
$x\ne 0,y\ne 0$ $\{\because AM\ge GM\}$
$\Rightarrow {(x+y)}^2\ge 4xy$

$\Rightarrow \frac{{(x+y)}^2}{4xy}\ge 1$
But, ${\cos }^2\theta =\frac{{(x+y)}^2}{4xy}\le 1$
$\therefore \frac{{(x+y)}^2}{4xy}=1\Rightarrow x=y$

Also, if $x+y=0$, then $se{c}^{2}x$ is not defined.
So, ans is $A$ and $B.$