Question

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

$x=y,x\ne 0$

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

$\Rightarrow \frac{4xy}{(x+y{)}^2}\ge 1$ $[\because se{c}^2\theta \ge 1]$

$\Rightarrow 4xy\ge (x+y{)}^2$

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

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

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

$\Rightarrow (x-y)\le 0$

$\Rightarrow x=y$

For $x=0,se{c}^2\theta$ will not be define,

$\Rightarrow x\ne 0$

$\therefore x=y$