MCQ
The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when
- ✓$x = y$
- B$x < y$
- C$x > y$
- DNone of these
${\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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