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 |
$\blacksquare$