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