Difference of Reciprocals of One Plus and Minus Sine
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Theorem
- $\ds \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x} = 2 \tan x \sec x$
Proof
| \(\ds \frac 1 {1 - \sin x} - \frac 1 {1 + \sin x}\) | \(=\) | \(\ds \frac {1 + \sin x} {1 - \sin^2 x} - \frac {1 - \sin x} {1 - \sin^2 x}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \sin x} {\cos^2 x}\) | Sum of Squares of Sine and Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \tan x} {\cos x}\) | Tangent is Sine divided by Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \tan x \sec x\) | Secant is Reciprocal of Cosine |
$\blacksquare$