Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » alt.algebra.help.independent

Topic: Proving sec^4 x - tan^4 x = sec^2 x + tan^2 x
Replies: 20   Last Post: Jan 29, 2006 1:48 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
ticbol

Posts: 116
Registered: 1/25/05
Re: Proving sec^4 x - tan^4 x = sec^2 x + tan^2 x
Posted: Jan 28, 2006 6:53 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


- wrote:
> Please help me in proving the following:
>
> sec^4 x - tan^4 x = sec^2 x + tan^2 x
>
> I only managed this far:
>
> (sec^2 x - tan^2 x)^2
> (sec^2 x - tan^2 x)(sec^2 x - tan^2 x)


----------------------------------------
sec^4(x) -tan^4(x) is not [sec^2(x) -tan^2(x)]^2
Like, a^4 -b^4 is not (a^2 -b^2)^2

a^2 -b^2 = (a+b)(a-b)
So,
a^4 -b^4 = (a^2 +b^2)(a^2 -b^2)
And so,
sec^4(x) -tan^4(x) = [sec^2(x) +tan^2(x)]*[sec^2(x) -tan^2(x)].
----------***

Then,
sin^2(x) +cos^2(x) = 1 --------trig identity.
Divide both sides by cos^2(X),
tan^2(x) +1 = sec^2(x)
Or,
1 = sec^2(x) -tan^2(x)
More correct,
sec^2(x) -tan^2(x) = 1 ------------------***

Now to the possible identity that you want to prove,
sec^4(x) -tan^4(x) = sec^2(x) +tan^2(x)

LHS =
= sec^4(x) -tan^4(x)
= [sec^2(x) +tan^2(x)]*[sec^2(x) -tan^2(x)]
= [sec^2(x) +tan^2(x)]*[1]
= [sec^2(x) +tan^2(x)]
= RHS

Proven.




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.