Question

${\sec }^2\theta =\frac{4xy}{(x+y{)}^2}$ is true, if

• A. $x=y$
• B. $x<y$
• C. $x>y$
• D. For every x and y

Answer 1

The correct option is A $x=y$

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

$\Rightarrow or{\cos }^2\theta =\frac{{(x+y)}^2}{4xy}$

Now, $0\le {\cos }^2\theta \le 1$

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

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

$\Rightarrow x^2+y^2+2xy=4xy$

$\Rightarrow x^2+y^2-2xy=0$

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

$\Rightarrow x-y=0$

$\Rightarrow x=y$

$\therefore$ If $x=y,$ then ${\sec }^2\theta =\frac{4xy}{{(x+y)}^2}.$

Hence, the answer is $x=y.$