sec
Question:

$\sec ^{2} x=\frac{4 x y}{(x+y)^{2}}$ is true if and only if

(a) x + y ≠ 0

(b) x = yx ≠ 0

(c) x = y

(d) x ≠0, y ≠ 0

Solution:

(b) x = yx ≠ 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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