sec^4y-tan^4y/sec^2y+tan^2y=1?

Question

im still having trouble with these...

Answer 1

(a^2-b^2)=(a-b)(a+b)
Hence,
sec^4y-tan^4y = (sec^2y-tan^2y)(sec^2y+tan^2y)

Substitute that back into the initial equation...

(sec^2y-tan^2y)(sec^2y+tan^2y)
------------------------------------------
(sec^2y+tan^2y)

and cancel common terms

= (sec^2y-tan^2y)
= 1 (a standard trig identity)

hope that solves it.

Answer 2

You need parentheses. It should say
(sec^4 y - tan^4 y)/(sec² y + tan² y) = 1

Then you factor the top as in Alg 1 to get
[ (sec² y - tan² y)(sec² y + tan² y) ] / (sec² y + tan² y)
and then you cancel out the common factor leaving
sec² y - tan² y = 1 which is an identity you've (presumably) memorized already.