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

 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

- wrote:
>
> 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.

Date Subject Author
1/28/06 -
1/28/06 William Elliot
1/28/06 Darrell
1/28/06 Darrell
1/28/06 Brian M. Scott
1/28/06 Darrell
1/29/06 Brian M. Scott
1/29/06 Stan Brown
1/29/06 Brian M. Scott
1/29/06 Brian M. Scott
1/29/06 Darrell
1/28/06 William Elliot
1/28/06 Brian M. Scott
1/28/06 Darrell
1/28/06 William Elliot
1/28/06 Darrell
1/28/06 Stan Brown
1/28/06 ticbol
1/28/06 Stan Brown