Answer
$1$
Work Step by Step
We know that $1+\tan^2 x=\sec^2 x $
$\dfrac{\sec^4 x -\tan^4 x}{\sec^2 x +\tan^2 x}=\dfrac{(\sec^2 x +\tan^2 x)(\sec^2 x -\tan^2 x)}{\sec^2 x +\tan^2 x}$
or, $=\sec^2 x -\tan^2 x$
or, $\dfrac{\sec^4 x -\tan^4 x}{\sec^2 x +\tan^2 x}=1$
