Question

Is the equation ${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}$ possible for real value of $x$ and $y?$
If $x=y=0$ Yes=1 No=3

Answer 1

Given, ${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}$
Since ${\sec }^2\theta \ge 1$, we get $\frac{4xy}{{(x+y)}^2}\ge 1$
or ${(x+y)}^2\le 4xy$
or ${(x+y)}^2-4xy\le 0$ or ${(x-y)}^2\le 0$
But for real values of $x$ and $y$, ${(x-y)}^2\ge 0$
Since ${(x-y)}^2=0,x=y$. Also $x+y\ne 0\Rightarrow x\ne 0,y\ne 0$
Therefor, The given equation ${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}$ is possible for real value of $x$ and $y$ only when $x=y(x\ne 0)$
Ans: 3