Tangent of Half Angle plus Quarter Pi
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Theorem
- $\map \tan {\dfrac x 2 + \dfrac \pi 4} = \tan x + \sec x$
Proof
Firstly, we have:
\(\text {(1)}: \quad\) | \(\ds \tan x\) | \(=\) | \(\ds \frac {2 \tan \frac x 2} {1 - \tan ^2 \frac x 2}\) | Double Angle Formula for Tangent |
Then:
\(\ds \map \tan {\frac x 2 + \frac \pi 4}\) | \(=\) | \(\ds \frac {\tan \frac x 2 + \tan \frac \pi 4} {1 - \tan \frac x 2 \tan \frac \pi 4}\) | Tangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan \frac x 2 + 1} {1 - \tan \frac x 2}\) | Tangent of $45 \degrees$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\tan \frac x 2 + 1} \paren {\tan \frac x 2 + 1} } {\paren {1 - \tan \frac x 2} \paren {\tan \frac x 2 + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan^2 \frac x 2 + 2 \tan \frac x 2 + 1} {1 - \tan^2 \frac x 2}\) | Difference of Two Squares, Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \tan \frac x 2} {1 - \tan^2 \frac x 2} + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tan x + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}\) | Double Angle Formula for Tangent: see $(1)$ above | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos^2 \frac x 2 - \sin^2 \frac x 2}\) | multiplying Denominator and Numerator by $\cos^2 \frac x 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos 2 \frac x 2}\) | Double Angle Formula for Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan x + \frac 1 {\cos x}\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \tan x + \sec x\) | Secant is Reciprocal of Cosine |
$\blacksquare$