Cosine over Sum of Secant and Tangent

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Theorem

$\displaystyle \frac {\cos x} {\sec x + \tan x} = 1 - \sin x$

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac {\cos x} {\sec x + \tan x}$	$=$	$\displaystyle \frac {\cos^2x} {1 + \sin x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Sum of Secant and Tangent
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {1 - \sin^2 x} {1 + \sin x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Sum of Squares of Sine and Cosine
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\left({1 + \sin x}\right) \left({1 - \sin x}\right)} {1 + \sin x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Difference of Two Squares
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle 1 - \sin x$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\cos x} {\sec x + \tan x}\)	\(=\)	\(\displaystyle \frac {\cos^2x} {1 + \sin x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Secant and Tangent
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {1 - \sin^2 x} {1 + \sin x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({1 + \sin x}\right) \left({1 - \sin x}\right)} {1 + \sin x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 - \sin x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)