Question

I need math help!!! ASAP!

Sec^2(-x) - tan^2(x) = ?

When simplified Pre-calc help!

Answer 1

hint.. sec^2(x) = 1/cos^2(-x)

Answer 2

tan^2(x) = sin^2x/cos^2x

Answer 3

I know, but I will try hold on.. I got answer = 1..

Answer 4

Let me work out..

Answer 5

I got 1 + tan2^x

How is that?

Answer 6

I got 1/cos^2(-x) - sin2^x/cos2^x and I( multiplied -sin^x/cos2^x by -1 to achieve a common denomonater of cos2^(-x).

So I put 1 + sin2^x/Cos2^-x = 1 -tan^2x?

Answer 7

?????????????????????

Answer 8

Do u hav the answer?

Answer 9

the answer 1 is correct!how did u get that?

Answer 10

No.

The answers are 1- 2tan^x , -1, 1 + 2tan^2x, 1. I picked one, but I just realized that is not true because to achieve one.. the first term would have to be simply sec^2x. not sec^2(-x).

Answer 11

ohh,now sec(-x)= sec x
thats an identity
so sec^2 (-x) = sec^2 (x)
thats why u got 1 as correct answer....

Answer 12

No!!! The identity is sec^2(-x) = -sec^2x.

Answer 13

ok,u know cos(-x)=cos x ??

Answer 14

Cos(-x) = -cosx.

Answer 15

Oh wait!! I just realied hatn!

Answer 16

Odd-even identities right?

Answer 17

can u please verify that for me again please....

Answer 18

Hartnn odd even identites right? That makes Sec(-x) = Secx.. which in turn will give me the correct answer I had which is one...

Answer 19

Right? It would become Sec^2(-x) = Sec^2x - tan^2x.

Answer 20

Which I transform into 1 + Tan^2x - Tan^2x = 1.

Answer 21

Correct?

Answer 22

i don't know whats odd even identities but,i am sure about
cos(-x)=cos x
and so sec(-x)=sec x

Answer 23

your rest of the work is correct...

Answer 24

Yes what you just stated is an Odd-Even Identity. Because cos and sec are even I believe. The rest are all odd functions making (Sin)(-x) = -sinx.. etc.. only cos and sec are even and become.. Cos(-x) = Cosx.. and Sec(-x)= Sec(x).

Answer 25

So I got the correct answer one?

Answer 26

thanks for telling me that :) now i know whats odd even identity
and yes u got correct answer as 1 :)

Answer 27

:D Thanks man! and anytime.. I am just trying to get to Calculus as a Sophomore.. I moved and had to redo the class this summer.. just wanted to verify that.. so I Can go submit all my answers :D

Answer 28

Thanks so much!