One Plus Tangent Half Angle over One Minus Tangent Half Angle
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Theorem
- $\dfrac {1 + \tan \frac x 2} {1 - \tan \frac x 2} = \sec x + \tan x$
Proof
\(\ds \frac {1 + \tan \frac x 2}{1 - \tan \frac x 2}\) | \(=\) | \(\ds \frac {1 + \frac {\sin \frac x 2}{\cos \frac x 2} }{1 - \frac {\sin \frac x 2}{\cos \frac x 2} }\) | Tangent is Sine divided by Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 - \sin \frac x 2}\) | multiply top and bottom by $\cos \dfrac x 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 - \sin \frac x 2} \times \frac {\cos \frac x 2 + \sin \frac x 2}{\cos \frac x 2 + \sin \frac x 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac{\cos^2 \frac x 2 + \sin^2 \frac x 2 + 2 \cos \frac x 2 \sin \frac x 2}{\cos^2 \frac x 2 - \sin^2 \frac x 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac{1 + 2 \cos \frac x 2 \sin \frac x 2}{\cos^2 \frac x 2 - \sin^2 \frac x 2}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac{1 + \sin x}{\cos x}\) | Double Angle Formula for Sine, Double Angle Formula for Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \sec x + \tan x\) | Sum of Secant and Tangent |
$\blacksquare$