sec

Solution:

(b) x = y, x ≠ 0

We hsve:

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

$\Rightarrow \frac{4 x y}{(x+y)^{2}} \geq 1 \quad\left[\because \sec ^{2} \mathrm{x} \geq 1\right]$

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

$\Rightarrow 4 x y \geq x^{2}+y^{2}+2 x y$

$\Rightarrow 2 x y \geq x^{2}+y^{2}$

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

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

$\Rightarrow x=y$

For $x=0, \sec ^{2} x$ will not be defined,

$\Rightarrow x \neq 0$

$\therefore x=y$