Secant Plus One over Secant Squared
Jump to navigation
Jump to search
Theorem
- $\dfrac {\sec x + 1} {\sec^2 x} = \dfrac {\sin^2 x} {\sec x - 1}$
Proof
\(\ds \frac {\sec x + 1} {\sec^2 x}\) | \(=\) | \(\ds \cos^2 x \paren {\frac 1 {\cos x} + 1}\) | Definition of Secant Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \cos x + \cos^2x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos x \paren {1 + \cos x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos x \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {1 - \cos x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - \cos^2 x} {\frac {1 - \cos x} {\cos x} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sin^2 x} {\frac 1 {\cos x} - 1}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sin^2 x} {\sec x - 1}\) | Definition of Secant Function |
$\blacksquare$