sec^4 y – tan^4 y

sec^4y-tan^4y=[sec^2y+tan^2y][sec^2y-tan^2y]=

[sec^2y+tan^2y][secy-tany][secy+tany] answer
or
[sec^2y + tan^2y][1-sin^2y]/cos^2y
[sec^2y+tan^2y]/cosy
[sec^2y+tan^2y]secy answer

Arthur
 
The answer I have in my book is just:

sec^2y+tan^2y

I wasnt sure on how to arrive at it.
 
Re: sec^4 y – tan^4 y

baselramjet said:
sec^4 y – tan^4 y
Click to expand...

You didn't tell us what the directions are. I hate to make assumptions, but I'll assume you are to simplify this expresssion.

Start by factoring as a difference of two squares:

(sec<SUP>2</SUP> y + tan<SUP>2</SUP> y) (sec<SUP>2</SUP> y - tan<SUP>2</SUP> y)

One of the fundamental identities tells us that
tan<SUP>2</SUP> y + 1 = sec<SUP>2</SUP> y
So,
1 = sec<SUP>2</SUP> y - tan<SUP>2</SUP> y

Substititute:

(sec<SUP>2</SUP> y + tan<SUP>2</SUP> y) ( 1)
or,
sec<SUP>2</SUP> y + tan<SUP>2</SUP> y You posted your textbook answer while I was typing this response. So....you can STOP HERE

Maybe you want this to be in terms of just one function? If so, you could substitute 1 + tan<SUP>2</SUP> y for sec<SUP>2</SUP> y:

(1 + tan<SUP>2</SUP> y) + tan<SUP>2</SUP> y
1 + 2 tan<SUP>2</SUP> y

If you perhaps need this to be in terms of sin and cos, you could take an additional step replacing tan<SUP>2</SUP> y with sin<SUP>2</SUP> y / cos<SUP>2</SUP> y.

It's a LOT easier for us to help you effectively if you include the directions for your problem AND show us what you have tried on your own.
 
sec^4y+tan^4y=[sec^2y+Tan^2y][sec^2y-tan^2y]

but sec^2y-tan^2y=1/cos^2y-sin^2y/cos^2y
= [1-sin^2y]/cos^2y
=cos^2y/cos^2y
=1

sec^4y-tan^2y=sec^2y+tan^2y answer in your book
Arthur
 
Re: sec^4 y – tan^4 y

Hello, baselramjet!

Simplify: \(\displaystyle \,\sec^4y\,-\,\tan^4y\)
Click to expand...

\(\displaystyle \sec^4y\,-\,\tan^4y\:=\:\underbrace{(\sec^2y\,-\,\tan^2y)}_{\text{This equals 1}}(\sec^2y\,+\,\tan^2y)\)

. . . . . . . . . . . \(\displaystyle =\;\sec^2y\,+\,\tan^2y\)

 
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