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Question

$sec to the power of 2 end exponent invisible function application theta equals fraction numerator 4 x y over denominator open parentheses x plus y close parentheses to the power of 2 end exponent end fraction$ is true if and only if

$x + y \neq 0$
$x = y, x \neq 0$
$x = y$
$x \neq 0, y \neq 0$

The correct answer is: $x equals y comma blank x not equal to 0$

Given, $sec to the power of 2 end exponent invisible function application theta equals fraction numerator 4 x y over denominator open parentheses x plus y close parentheses to the power of 2 end exponent end fraction$
Now, $sec to the power of 2 end exponent invisible function application theta greater or equal than 1 rightwards double arrow fraction numerator 4 x y over denominator open parentheses x plus y close parentheses to the power of 2 end exponent end fraction greater or equal than 1$
$rightwards double arrow open parentheses x plus y close parentheses to the power of 2 end exponent less or equal than 4 x y$
$rightwards double arrow open parentheses x plus y close parentheses to the power of 2 end exponent minus 4 x y less or equal than 0$
$rightwards double arrow open parentheses x minus y close parentheses to the power of 2 end exponent less or equal than 0$
But for real values of $x$ and $y$ , $open parentheses x minus y close parentheses to the power of 2 end exponent greater or equal than 0 blank rightwards double arrow open parentheses x minus y close parentheses to the power of 2 end exponent equals 0$
$therefore x equals y$
Also $x plus y not equal to 0 rightwards double arrow x not equal to 0. y not equal to 0$

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