Tangent over Secant Plus One

Theorem

$\displaystyle \frac {\tan x} {\sec x + 1} = \frac {\sec x - 1} {\tan x}$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {\tan x} {\sec x + 1}\)	\(=\)	\(\displaystyle \frac {\sin x} {\cos x \left({\frac 1 {\cos x} + 1}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By definition of tangent and secant
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\sin x} {1 + \cos x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\sin^2 x} {\sin x \left({1 + \cos x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Multiply by $1 = \dfrac {\sin x} {\sin x}$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {1 - \cos^2 x} {\sin x \left({1 + \cos x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum of Squares of Sine and Cosine
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({1 + \cos x}\right) \left({1 - \cos x}\right)} {\sin x \left({1 + \cos x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {1 - \cos x} {\sin x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\cos x \left({\frac 1 {\cos x} - 1}\right)} {\sin x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\sec x - 1} {\tan x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By definition of tangent and secant

$\blacksquare$