Tangent over Secant Plus One

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Theorem

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


Proof

\(\ds \frac {\tan x} {\sec x + 1}\) \(=\) \(\ds \frac {\sin x} {\cos x \paren {\sec x + 1} }\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {\sin x} {\cos x \paren {\frac 1 {\cos x} + 1} }\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \frac {\sin x} {1 + \cos x}\) simplifying
\(\ds \) \(=\) \(\ds \frac {\sin^2 x} {\sin x \paren {1 + \cos x} }\) Multiply by $1 = \dfrac {\sin x} {\sin x}$
\(\ds \) \(=\) \(\ds \frac {1 - \cos^2 x} {\sin x \paren {1 + \cos x} }\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds \frac {\paren {1 + \cos x} \paren {1 - \cos x} } {\sin x \paren {1 + \cos x} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac {1 - \cos x} {\sin x}\)
\(\ds \) \(=\) \(\ds \frac {\cos x \paren {\frac 1 {\cos x} - 1} } {\sin x}\)
\(\ds \) \(=\) \(\ds \frac {\cos x \paren {\sec x - 1} } {\sin x}\) Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \frac {\sec x - 1} {\tan x}\) Tangent is Sine divided by Cosine

$\blacksquare$