Secant Plus One over Secant Squared

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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          

$\blacksquare$

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