Posted by olivia on .
secx/secxtanx=sec^2x+ secx+tanx

precal PLEASE HELP! 
Damon,
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 ]/ [1sin^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,
Sorry, it was a typo, it's supposed to be:
sec/(secxtanx)=sec^2x+secxtanx 
precal PLEASE HELP! 
Reiny,
LS = (1/cosx)/(1/cosx  sinx/cosx)
= (1/cosx)/( (1sinx)/cox )
= (1/cosx)(cosx)/(1sinx)
= 1/(1sinx)
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/(1sinx) * (1+sinx)/(1+sinx)
= (1+sinx)(1  sin^2 x)
= (1+sinx)/cos^2 x
= RS 
precal PLEASE HELP! 
Damon,
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] 
precal PLEASE HELP! 
olivia,
:)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/sinx1/(sinx))? 
precal PLEASE HELP! 
Damon,
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 
precal PLEASE HELP! 
olivia,
thanks alot!!!!!!!!!!!!!!