Question:
$\sec ^{2} x=\frac{4 x y}{(x+y)^{2}}$ is true if and only if
(a) x + y ≠ 0
(b) x = y, x ≠ 0
(c) x = y
(d) x ≠0, y ≠ 0
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,
$\Rightarrow x \neq 0$
$\therefore x=y$
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