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

\(\ds \)

\(=\)

\(\ds 1 - \sin x\)

$\blacksquare$