સમીકરણ {sec ^2}theta = frac{{4xy}}{{{{( | vedclass

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Question

(a) Since ${\cos ^2}	heta \le 1$

${\sec ^2}	heta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $

$\Rightarrow 4xy \ge {(x + y)^2} $

$\Rightarrow {(

Vedclass · Accepted Answer

$x = y$

Vedclass · Answer

(a) Since ${\cos ^2}	heta \le 1$

${\sec ^2}	heta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $

$\Rightarrow 4xy \ge {(x + y)^2} $

$\Rightarrow {(x - y)^2} \le 0$

It is possible only when $x = y$, .$(x,\,y \in R)$

સમીકરણ ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ તો જ શક્ય છે જો . . . .

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