If ABCD is a quadrilateral, then ${\sec }^2\left(\frac{A+B}{4}\right)-{\cot }^2\left(\frac{C+D}{4}\right)=$

${\sec }^2\left[\frac{A+B}{4}\right]-{\cot }^2\left[\frac{C+D}{4}\right]$

In a quadrilateral each angle is $90°$

$={\sec }^2\left[\frac{180°}{4}\right]-{\cot }^2\left[\frac{180°}{4}\right]$

$={\sec }^2(45°)-{\cot }^2(45°)$

$={\left(\sqrt{2}\right)}^2-{\left(1\right)}^2$

$=2-1$

$=1$