Difference of Reciprocals of One Plus and Minus Sine

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Theorem

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

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}$	$=$	$\displaystyle \frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Difference of Two Squares
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {2 \sin x} {\cos^2 x}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Sum of Squares of Sine and Cosine
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle 2 \tan x \ \sec x$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by definition of tangent and secant

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}\)	\(=\)	\(\displaystyle \frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2 \sin x} {\cos^2 x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \tan x \ \sec x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of tangent and secant