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June 24, 2016
Posted by **olivia** on Sunday, March 25, 2012 at 9:21pm.

- precal- PLEASE HELP! -
**Damon**, Sunday, March 25, 2012 at 9:49pmsuspect 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 - precal- PLEASE HELP! -
**olivia**, Sunday, March 25, 2012 at 9:58pmSorry, it was a typo, it's supposed to be:

sec/(secx-tanx)=sec^2x+secxtanx - precal- PLEASE HELP! -
**Reiny**, Sunday, March 25, 2012 at 10:16pmLS = (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 - precal- PLEASE HELP! -
**Damon**, Sunday, March 25, 2012 at 10:18pmWell 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] - precal- PLEASE HELP! -
**olivia**, Sunday, March 25, 2012 at 10:27pm:)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))? - precal- PLEASE HELP! -
**Damon**, Monday, March 26, 2012 at 4:37amIf 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 - precal- PLEASE HELP! -
**olivia**, Monday, March 26, 2012 at 9:54amthanks alot!!!!!!!!!!!!!!