Trig Functions question

mjbc

1. The problem statement, all variables and given/known data

Simplify: cosx / (1+sinx)

2. The attempt at a solution

1 / (sec x)(1+sin x)

1 / (sec x + (sin x / cos x))

1 / (sec x + tan x)

I know that the answer is sec x - tan x but don't know how to get there.

Any help would be appreciated.

jing

Do you know what the value of [tex]sec^2x-tan^2x[/tex] is?

The difference of two squares helps too.

mjbc

sec^2 (x) - tan^2 (x) = 1

but nothing's squared in the problem

jing

Not until you do some multiplying.

Do you understand what I mean by the difference of two squares?

Dick

You could also just multiply top and bottom of your given expression by (1-sin(x)) and expand. It's really the same thing as jing is suggesting.

mjbc

OK, So it's

(sec x - tan x) / (sec x + tan x)(sec x -tan x)

= (sec x - tan x) / 1

= sec x -tan x

Thank you so much!!