Question

(sec^2x-1)/sec^2X

Answer 1

IT SAYS SIMPLIFY

Answer 2

ok, do what?

Answer 3

here is what I have I replaced sec^2x= 1-cosx

Answer 4

\[\frac{ \sec^2 x -1 }{ \sec^2x }= 1- \frac{ 1 }{ \sec^2x }= 1- \cos^2x= \sin^2 x\]

Answer 5

just one line

Answer 6

is that it?

Answer 7

yeap! break the numerator into 2 parts. and manipulate until you cannot. then combine them together again

Answer 8

let me redo it

Answer 9

ok take your time

Answer 10

how did you come from 1-1/sin^2x in the denominator

Answer 11

that 1-cos^2x? isn't that supposed to be in the denominator?

Answer 12

\[\frac{ a+b }{ c }= \frac{ a }{ c }+\frac{ b }{ c }\]

Answer 13

you have a disease of turning everything to sin and cos, I don't, I manipulate right on sec^2

Answer 14

I know but I replaced sec^2x by 1-cos^2x

Answer 15

I know , Sydney gave you that method, but it's no need to do all the time

Answer 16

is 1/sec^2 = cos^2?

Answer 17

yes

Answer 18

can you show me how?

Answer 19

sec = 1/cos ---> cos = 1/sec
cos^2 = 1/sec^2

Answer 20

Ok thanks. are you going to be online for awhile? I will try the others now

Answer 21

sure, post other