The equation ^2 = 4xy (x + y) ^2 is only possible when - Vidyadip

MCQ

The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when

✓
$x = y$
B
$x < y$
C
$x > y$
D
None of these

✓

Answer

Correct option: A.

$x = y$

a
(a) Since ${\cos ^2}\theta \le 1$

${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $

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

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

It is possible only when $x = y$, .$(x,\,y \in R)$

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