Question

${\sec }^2\theta =\frac{4xy}{{\left(x-y\right)}^2},Wherex\in R,y\in R$ is true if and only if.

• A. $x+y\ne 0$
• B. $x+y,x\ne 0$
• C. $x=y$
• D. none

Answer 1

The correct option is A $x+y\ne 0$

Solution :

Given $se{c}^2\theta =\frac{4xy}{(x+y{)}^2}$ , we know $se{c}^2\theta \ge 1$

$\therefore \frac{4xy}{(x+y{)}^2}\ge 1$

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

$\Rightarrow x^2+y^2-2xy\le 0$

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

$\Rightarrow x=y$