Question

verify: cosA/secAsinA=cscA-sinA

Answer 1

why rnt u geting it right??????

Answer 2

i have to make one side look like the other side

Answer 3

are they really equal o.O

Answer 4

thats what it says on my paper o-o'

Answer 5

\[\frac{\cos A}{\frac{1}{\cos A}\sin A}=\frac{1}{\sin A}-\sin A\]

Answer 6

but cosA/sinA/cosA = 1/sinA @Mertsj o.O

Answer 7

i got as far as Mertsj has it but i always get confused on what to do after that

Answer 8

ohh i got it...

Answer 9

\[\frac{\cos A}{\frac{\sin A}{\cos A}}\] \[\frac{\frac{\cos A}{\cos A}}{\frac{\sin A}{\cos^2 A}}\] i divided numerator and denom by cosA

\[\frac{1}{\frac{\sin A}{\cos^2 A}}\] \[\frac{\cos^2 A}{\sin A}\] \[\frac{1 - \sin^2 A}{\sin A}\] canyou take over from here?

Answer 10

you have to do some magic...it's not too straightforward

Answer 11

i got lost with the first part o.o'

Answer 12

first part...sec = 1/cos so cos/sec sin = cos/(sin/cos)

Answer 13

oooh ok ^^'

Answer 14

so do you get the succeeding steps?

Answer 15

yes except for the second to last one

Answer 16

1/sin/cos^2

complex frac

1/1 * cos^2/sin

Answer 17

ok ^^ i got it from here

Answer 18

no wait on the last part do i multiply again?

Answer 19

last part? it's an identity

sin^2 + cos^2 = 1

therefore cos^2 = 1 - sin^2

Answer 20

\[\frac{\cos ^2A}{\sin A}=\frac{1- \sin ^2A}{\sin A}=\frac{1}{\sin A}-\frac{\sin ^2A}{\sin A}=\csc A-\sin A\]

Answer 21

oh ok i thought i did it wrong

Answer 22

thx ^^