Answer
$\frac{sec^4y-tan^4y}{sec^2y+tan^2y}=1$
The equation is true.
Work Step by Step
$\frac{\sec^4y-\tan^4y}{\sec^2y+\tan^2y}=1$
Factor the numerator, so
$\sec^4y - \tan^4y = (\sec^2y+\tan^2y)\times(\sec^2y-\tan^2y)$
where $\sec^2y-\tan^2y = 1$
$\frac{(\sec^2y+\tan^2y)\times(\sec^2y-\tan^2y)}{\sec^2y+\tan^2y}= \frac{(\sec^2y+\tan^2y)\times1}{\sec^2y+\tan^2y}=1$
