Question

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

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

Answer 1

Answer: (C)

$x=y$

${sec}^2\theta =\frac{4xy}{{(x+y)}^2}$

is true only if $4xy\ge {(x+y)}^2$

or $4xy\ge x^2+2xy+y^2$

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

$\Rightarrow x=y$ [$\because$ ${(x-y)}^2<0$ is not possible]

[C]