Question

Quick question:

Does sec^4(x)-tan^4(x) over sec^2(x)+tan^2(x) simplify into cos^2(x) over cos^2(x), which equals 1?

Answer 1

WAIT never mind, it's wrong, isn't it?

Answer 2

I think the 1 is correct.

I factored the numerator (difference of two squares) and divided out the denominator of sec² x + tan²x. That left sec² x - tan²x. Then, I used the identity: 1 + tan²x = sec² x to get 1. Check that.

Answer 3

Could you show me each step you did? I'm still a bit confused

Answer 4

Factor the numerator:

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

Answer 5

OHHH okay got it. Thank you so much! i understand now.

Answer 6

Do you see that the denominator of sec^2(x)+tan^2(x) is a factor of the numerator?

Answer 7

Yes.

Answer 8

After division, you are left with this: sec^2(x) - tan^2(x)

Answer 9

Okay.

Answer 10

Trig Identity: 1 + tan^2(x) = sec^2(x).

Learn those identities.

Answer 11

Alright!

Answer 12

Take the numerator of sec^2(x) - tan^2(x) and in place of sec^2(x), put 1 + tan^2 (x) to get ....

Answer 13

sec^2(x), correct?

Answer 14

1 + tan^2 (x) - tan^2x = 1.

Answer 15

sec^2(x) - tan^2(x) = (1 + tan^2 (x) ) - tan^2x = 1+ tan^2 (x) - tan^2(x) = 1.

Answer 16

I understand now. Thank you so much!

Answer 17

You are welcome.