Sum of Reciprocals of One Plus and Minus Sine

From ProofWiki

Jump to: navigation, search

Theorem

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

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

$\blacksquare$

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