Question

(cosx-1/sinx)- sinx/ cosx-1

Answer 1

\[\frac{ \cos x-1 }{ sinx }-\frac{ sinx }{ cosx -1 }\]

Answer 2

combine everything into one fraction.

Answer 3

make them have the same denominator

Answer 4

\[\frac{ (cosx -1)^2 -\sin^2x }{ sinx (cosx-1) }\]

Answer 5

(cosx-1)(cosx-1)(/sinx(cosx-1)- sinx(sinx)/ cosx-1(sinx)

Answer 6

combine what?

Answer 7

got it? or need more explanation? I ll come up on another way.

Answer 8

now, (a - b ) (a - b) = (a - b)^2
yours (cosx -1)(cosx -1) = (cosx -1 )^2
got it?

Answer 9

so, I just combine to have only 1 fraction.
I can do it because I have the same denominator already (the first step I did)
got it?

Answer 10

if you cannot, don't try. just follow me and say yes/no if you get or not.

Answer 11

so, are you with me now?

Answer 12

continue. you have (cos x-1)^2 - sin^2x at numerator, right? open the square. do it at the paper, I do it here, you just compare ours

Answer 13

\[\frac{ \cos^2x -2cosx + 1-\sin^2x}{sinx (cosx -1) }\]

Answer 14

replace 1 = sin^2x + cos^2 x

Answer 15

you have

Answer 16

\[\frac{ \cos^2x-2cosx+\sin^2x+\cos^2x-\sin^2x }{ sinx(cosx-1)}\]

Answer 17

from the sin^2x +cos^2x=1 but this one is negative

Answer 18

cancel out the sin^2

Answer 19

you mean replace it with 1-cos^x?

Answer 20

replace 1 only, see mine, again

Answer 21

what did you replaced it with? I though 1-cos^2x

Answer 22

It is (cos x -1)^2 not yours

Answer 23

yes but I didn't see it in your answer

Answer 24

no I wrote it according to the formula that says: 1-cos^2x= sin^2x

Answer 25

i didn't give out the answer, we are on the way

Answer 26

i meant i replaced it with that

Answer 27

I make the stuff base on what you post. I don't care and I don't know where do you you pick it up.

Answer 28

i know but i don't get what you had on top since I replaced sin^2x by 1-cos^2x

Answer 29

i don't replace sin^2 x . It's there still. I just combine terms together

Answer 30

\[\frac{ 2\cos^2x -2cosx }{ sinx(cosx-1) }= \frac{ 2cosx(cosx-1) }{sinx( cosx -1) }\]
= \[\huge\frac{ 2cosx }{ sinx}= 2 cot x\]

Answer 31

got it?

Answer 32

give me a sec.

Answer 33

no I am not but let me go over it again

Answer 34

Madina \[\sin^2x + \cos^2 x =1\]

Answer 35

This is GREAT i guess i can do whatever works. My moral is up again. Thanks.

Answer 36

ok, I have to say, thanks god, too

Answer 37

lol

Answer 38

the part I cancel out the (cosx-1)?

Answer 39

just cancel, numerator and denominator, cancel out , done