Reciprocal of One Minus Secant

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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          

$\blacksquare$

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