Showing $(1 + \sin x)^{-1} = \sec^2 x - \sec x \tan x$

Question

I was asked to show the following:

$$\frac {1}{1 + \sin x} = \sec^2 x - \sec x \tan x$$

I proceeded to factorize and simplify..

$\sec x (\sec x - \tan x)$$

$\sec x (\frac {1}{\cos x} - \frac {\sin x}{\cos x}) $

$\sec x (\frac {1 - \sin x}{\cos x}) $

$\frac {1 - \sin x}{\cos ^ 2 x}$

$\frac {1 - \sin x}{1 - \sin ^ 2 x} $

and then i'm lost..

Any help will be much appreciated!

Answer 1

You are almost done. The denominator can be rewritten as $(1-\sin{x})(1+\sin{x})$.