sec^2A≥ 1 4xy(x+y)^2≥ 1 #160; 4xy≥ (x+y)^2 #160; (x+y)^2≤ 4xy #160; x^2+y^2+2xy≤ 4xy #160; x^2+y^2 2xy≤ 0 (x y)^2≤ 0 ...

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Inverse of a Function

If 2θ=4xyx+...

Question

If ${sec}^{2} θ = \frac{4 x y}{(x + y)^{2}}$ then.

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Solution

$s e c^{2} A \geq 1$

$\frac{4 x y}{(x + y)^{2}} \geq 1$

$4 x y \geq (x + y)^{2}$
$(x + y)^{2} \leq 4 x y$
$x^{2} + y^{2} + 2 x y \leq 4 x y$
$x^{2} + y^{2} - 2 x y \leq 0$
$(x - y)^{2} \leq 0$

Since square of any number cannot be negative the only case possible is ,
$(x - y)^{2} = 0$
$x - y = 0$
$x = y$

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