Expert: Steve Holleran - 3/10/2007

Question
Hi,
I have been trying to figure this problem out... how do i prove this identity?

sec^6x-tan^6x=1+3tan^2xsec^2x

Answer
Hi Britney,

This is a real winner.  The only way I could get this to work out is to first use the factoring for a difference of cubes, and then use a real sneaky "trick" to do some other factoring.

Okay, to get started , we've got to think of sec^6 x as
(sec^2 x)^3 and tan^6 x as (tan^2 x)^3.  To make the notation easier to type, I'm going to let A = sec^2 x and B = tan^2 x.  So then we have this:

sec^6 x - tan^6 x = A^3 - B^3

                 = (A - B)(A^2 + AB + B^2)

The first parenthesis here, (A -B), stands for sec^2 x - tan^2 x , which by identity = 1.  So what we really have now is just A^2 + AB + B^2.

Okay, now for the trick:  Subtract, then add, 3AB:

  A^2 + AB -3AB + B^2  + 3AB  (So you've really added 0 )

Now you have :

  A^2 - 2AB + B^2  + 3AB  .   If you group the first three terms together, it will factor as (A - B)^2:

 [A^2 - 2AB + B^2]  + 3AB

= [A - B]^2  + 3AB.  Now replace A and B

= [sec^2 x - tan^2 x]^2  + 3 sec^2 x tan^2 x

Once again, the stuff in the brackets = 1 by identity, so you now have

     1^2  + 3 sec^2 x tan^2 x   and this is your result.

I hope you were able to follow this ok.  I just thought the substitutions would be easier to read than all the trig functions and exponents.

Steve Holleran

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