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

$\blacksquare$