Reciprocal of One Plus Cosecant

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Theorem

$\displaystyle \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2} = \frac 1 {1 + \csc x}$


Proof

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2}\) \(=\) \(\displaystyle \frac {1 - \sin^2 x + 3 \sin x - 1} {1 - \sin^2 x + 2 \sin x + 2}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Sum of Squares of Sine and Cosine          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac {\sin^2 x - 3 \sin x} {\sin^2 x - 2 \sin x - 3}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac{\sin x \left({\sin x - 3}\right)} {\left({\sin x - 3}\right) \left({\sin x + 1}\right)}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac {\sin x} {\sin x + 1}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + \dfrac 1 {\sin x} }\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1 + \csc x}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          by definition of cosecant          

$\blacksquare$

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