Question

If $se{c}^2\theta =\frac{4xy}{(x+y{)}^2}$ is true, then which of the following is true? $(x\ne -y)$

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

Answer 1

The correct option is B $x=y,x\ne 0$

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

We know, $se{c}^2\theta \ge 1$
$\frac{4xy}{(x+y{)}^2}\ge 1$
$4xy\ge (x+y{)}^2$
$x^2-2xy+y^2\le 0$
$(x-y{)}^2\le 0$

Now, $(x-y{)}^2<0$ is not possible
$\therefore x-y=0$
$\Rightarrow x=y$ and $x\ne 0$