$(sec{x}^{\circ }-tan{x}^{\circ })(sec{x}^{\circ }+tan{x}^{\circ })$

Let the given triangle be ABC, where $\angle ABC={90}^{\circ }$, $AB=1,BC=y,AC=2$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$2^2=1^2+y^2$
$y^2=3$
$y=\sqrt{3}$
Now, $(\sec x^0-\tan x^0)(\sec x^0+\tan x^0)$
= ${\sec }^2{x}^{\circ }-{\tan }^2{x}^{\circ }$
= $(\frac{H}{B}{)}^2-(\frac{P}{B}{)}^2$
= $(\frac{2}{1}{)}^2-(\frac{\sqrt{3}}{1}{)}^2$
= $4-3$
= $1$