secx/secx-tanx=sec^2x+ secx+tanx

Answer 1

suspect you might mean

sec x / (sec x -tan x) = sec^2 x + sec x + tan x
work on the left
1/cos x / [ 1/cos x -sin x/cos x ] =

1/[1 - sin x] =
[ 1 + sin x ]/ [1-sin^2 x] =
[1+sin x ]/cos^2 x =

work on the right
= 1/cos^2 x + 1/cos x + sin x/cos x)
= (1/cos^2 x) [1 + cos x + sin x cos x]
sorry, I can not make that come out. Please check for typos

Answer 2

Sorry, it was a typo, it's supposed to be:
sec/(secx-tanx)=sec^2x+secxtanx

Answer 3

LS = (1/cosx)/(1/cosx - sinx/cosx)
= (1/cosx)/( (1-sinx)/cox )
= (1/cosx)(cosx)/(1-sinx)
= 1/(1-sinx)

RS = 1/cos^2 x + (1/cosx)(sinx/cosx)
= 1/cos^2 x + sinx/cos^2 x
= (1+ sinx)/cos^2x

back to LS
LS = 1/(1-sinx) * (1+sinx)/(1+sinx)
= (1+sinx)(1 - sin^2 x)
= (1+sinx)/cos^2 x
= RS

Answer 4

Well then we are finished
work on the right
= 1/cos^2 x + (1/cos x * sin x/cos x)
= (1/cos^2 x) [1 + sin x]

Answer 5

:)thanks!!! I've been trying to figure this out for a while now, can you explain this to me: what do I do when i get to something like this
(cosx/sinx-1/(sinx))?

Answer 6

If you see (sin x - 1) on the bottom try multiplying top and bottom by (sin x + 1)
like
1/(sin x - 1) * [ (sin x + 1)/(sin x+1) ]

= (sin x + 1) / (sin^2 x - 1)

= (sin x +1)/-cos^2 x

Answer 7

thanks alot!!!!!!!!!!!!!!