Reciprocal of One Minus Secant
From ProofWiki
Theorem
- $\displaystyle \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3} = \frac 1 {1 - \sec x}$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {\sin^2 x + 2 \cos x - 1} {\sin^2 x + 3 \cos x - 3}\) | \(=\) | \(\displaystyle \frac {1 - \cos^2 x - 2 \cos x - 1} {1 - \cos^2 x + 3 \cos x - 3}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Sum of Squares of Sine and Cosine | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\cos^2 x - 2 \cos x} {\cos^2 x - 3 \cos x + 2}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\cos x \left({\cos x - 2}\right)} {\left({\cos x - 1}\right) \left({\cos x - 2}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\cos x} {\cos x - 1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {1 - \frac 1 {\cos x} }\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 {1 - \sec x}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by definition of secant |
$\blacksquare$