Cosine over Sum of Secant and Tangent

Theorem

$\dfrac {\cos x} {\sec x + \tan x} = 1 - \sin x$


Proof

\(\ds \frac {\cos x} {\sec x + \tan x}\) \(=\) \(\ds \frac {\cos^2 x} {1 + \sin x}\) Sum of Secant and Tangent
\(\ds \) \(=\) \(\ds \frac {1 - \sin^2 x} {1 + \sin x}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \frac {\paren {1 + \sin x} \paren {1 - \sin x} } {1 + \sin x}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 1 - \sin x\)

$\blacksquare$

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