Question

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

• A. $x=y$
• B. $x<y$
• C. $x>y$
• D. None of these

Answer 1

The correct option is A $x=y$

We have

$se{c}^2\theta =\frac{4xy}{(x+y{)}^2}$

$co{s}^2\theta =\frac{(x+y{)}^2}{4xy}$

We knw that

$0\le co{s}^2\theta \le 1$

$0\le \frac{(x+y{)}^2}{4xy}\le 1$

$0\le (x+y{)}^2\le 4xy$

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

$0\le x-y\le 0$

$x=y$