Question

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

The correct option is B

$x=y,x\ne 0$

Explanation for the correct answer:

Finding the condition:

Given,
${\text{sec}}^2\theta =\frac{4xy}{{\left(x+y\right)}^2}$
We know that; ${\sec }^2\theta \ge 1$
Therefore, $\frac{4\mathrm{xy}}{{\left(\mathrm{x}+\mathrm{y}\right)}^2}\ge 1$

Now, taking the numerator alone
$4\mathrm{xy}\ge {(\mathrm{x}+\mathrm{y})}^2$
${\Rightarrow (\mathrm{x}+\mathrm{y})}^2–4\mathrm{xy}\le 0$
$\Rightarrow {(\mathrm{x}–\mathrm{y})}^2\le 0$
$\Rightarrow x-y=0$
$\Rightarrow x=y$
Now, taking the denominator alone
$x+y\ne 0$ [denominator cannot be $0$]
So, $2x\ne 0$ [substituting $x=y$]
Therefore $x\ne 0$

Thus, we can conclude $x=y,x\ne 0$

Hence, option (B) is the correct answer.