Reciprocal of One Minus Secant
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Theorem
- $\dfrac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3} = \dfrac 1 {1 - \sec x}$
Proof
\(\ds \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3}\) | \(=\) | \(\ds \frac {1 - \cos^2 x - 2 \cos x - 1} {1 - \cos^2 x + 3 \cos x - 3}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos^2 x - 2 \cos x} {\cos^2 x - 3 \cos x + 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos x \paren {\cos x - 2} } {\paren {\cos x - 1} \paren {\cos x - 2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos x} {\cos x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 - \frac 1 {\cos x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {1 - \sec x}\) | Definition of Secant Function |
$\blacksquare$