Please help me prove this trig identity (please include justification) (sinA^4 - cosA^4) secA^2 =secA^2-2
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Originally Posted by danny88 (sinA^4 - cosA^4) secA^2 =secA^2-2 Are you sure that that's not supposed to be
If that is what you are trying to prove, begin by factoring the as a difference of squares. Then use the Pythagorean identity and the rest should be obvious.
I would really appreciate if you would show me step by step because I seem to be making an error in my steps.
Originally Posted by danny88 I would really appreciate if you would show me step by step because I seem to be making an error in my steps. Well, how about you post the work that you have done so that I can see where you went wrong?
(sin^2 A+cos^2 A)(sin^2A-cos^2A) * sec^2A
(1)*(sin^2A-cos^2A) * sec^2A
1*(sinA-cosA)^2 * sec^2A
Thank you. Originally Posted by danny88 1*(sinA-cosA)^2 * sec^2A This is your error. We cannot distribute powers over addition or subtraction and, likewise, we cannot "factor out" a power like that. is not necessarily equal to . Instead, convert the secants to cosines:
Can you continue from here?
sin^2 A/ cos^2A =tan^2A?
then -cos^2A/cos^2A= -1 ?
Originally Posted by danny88 sin^2 A/ cos^2A =tan^2A?
then -cos^2A/cos^2A= -1 ? Correct. So we have
Is there an identity we can use that relates with
but how does that work out?
Originally Posted by danny88 1= sec^2A-tan^2A
but how does that work out? Solve for
Now substitute that into our expression:
thanks a lot of the help, I really appreciate it.
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