Thread: Math Trig Identity

Please help me prove this trig identity (please include justification)

(sinA^4 - cosA^4) secA^2 =secA^2-2

appreciated

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
1*(1-2sinAcosA)* 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

1= sec^2A-tan^2A
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.