Question

prove using identities that ((sec^2)x)/(((sec^2)x)-1)=(csc^2)x

Answer 1

Okay, so let me try this:
did you mean:
\[Sec^2(x)/(Sec^2x - 1) = Csc^2(x)\]

Answer 2

yea. sorry i don't know how get it like that on my computer

Answer 3

okay so now workin on it

Answer 4

thank you so much

Answer 5

There are a few Sec and Csc identities, but I never remember them so im just gonna convert em to Sin and Cos:
Sec(x) = 1/Cos(x)
Csc(x) = 1/Sin(x)

Answer 6

\[[1/Cos^2x] / [ (1/Cos^2x)-1 ] = 1/Sin^2x\]

Answer 7

Now let's simplify this part:
\[(1/Cos^2x) - 1\] = \[(1-Cos^2x)/Cos^2x\]

Answer 8

wait. how did you do that?

Answer 9

find the common denominator

Answer 10

1 = Cos^2 / Cos^2... get it?

Answer 11

oh . yea, sorry. i got it

Answer 12

k

Answer 13

So now you have to divide 1/Cos^2x by (1-Cos^2x)/Cos^2x which is the same as multiplying by its inverse

Answer 14

\[[1/Cos^2x] * [Cos^2x/(1-Cos^2x)] = 1/Sin^2x\]

Answer 15

Cancel out cos^2s

Answer 16

\[1/(1-Cos^2x) = 1/Sin^2x\]

Answer 17

Cross multiply:

Answer 18

wait. can't you just say 1 / (sin^2x) =csc^2x

Answer 19

and then be done? or is that skipping a step?

Answer 20

yeah, but i prefer keeping everything in sin and cosine

Answer 21

bcuz its easy to simplify

Answer 22

\[Sin^2x = 1 - Cos^2x\]

Answer 23

See that's proved. Sin^2x + Cos^2x = 1!!!!

Answer 24

get it?

Answer 25

yes. thank you so much!

Answer 26

k off to helpin someone else.. any questions:
email me bahrom.cfo@gmail.com and if u liked my explanation, bcum a fan YAY!

Answer 27

deff will. that was so helpful. thanks again!