Reciprocal of One Plus Cosecant

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Theorem

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


Proof

\(\ds \frac {\cos^2 x + 3 \sin x - 1} {\cos^2 x + 2 \sin x + 2}\) \(=\) \(\ds \frac {1 - \sin^2 x + 3 \sin x - 1} {1 - \sin^2 x + 2 \sin x + 2}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \frac {\sin^2 x - 3 \sin x} {\sin^2 x - 2 \sin x - 3}\)
\(\ds \) \(=\) \(\ds \frac {\sin x \paren {\sin x - 3} } {\paren {\sin x - 3} \paren {\sin x + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\sin x} {\sin x + 1}\)
\(\ds \) \(=\) \(\ds \frac 1 {1 + \dfrac 1 {\sin x} }\)
\(\ds \) \(=\) \(\ds \frac 1 {1 + \csc x}\) Definition of Real Cosecant Function

$\blacksquare$