Question

Is ${\sec }^2\theta =\frac{4xy}{(x+y{)}^2}$ possible for real value of $x$ and $y$?

Answer 1

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

$\frac{1}{\cos \theta }=\sqrt{\frac{4xy}{(x+y{)}^2}}$

$\frac{1}{\cos \theta }=\frac{2\sqrt{xy}}{x+y}$

$\cos \theta =\frac{x+y}{2\sqrt{xy}}$

We know $\cos \theta$ lies between $-1$ & $1$

$-1\le \frac{x+y}{2\sqrt{xy}}\le 1$ …….$\left(1\right)$

$AM\ge Gm$

for x & y $\Rightarrow \frac{x+y}{2}\ge \sqrt{xy}$

$\frac{x+y}{2\sqrt{xy}}\ge 1$ ………$\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$\frac{x+y}{2\sqrt{xy}}=1$

$(x+y{)}^2=4xy$

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

$\therefore x=y$

$\therefore$ Infinite solutions. $[x\pm 0,y\pm 0]$.