Tangent of i/Proof 1

Theorem

$\tan i = \paren {\dfrac {e^2 - 1} {e^2 + 1} } i$

Proof

\(\ds \tan i\)

\(=\)

\(\ds \frac {\sin i} {\cos i}\)

Definition of Complex Tangent Function

\(\ds \)

\(=\)

\(\ds \frac {\paren {\frac e 2 - \frac 1 {2 e} } i} {\frac e 2 + \frac 1 {2 e} }\)

Sine of $i$ and Cosine of $i$

\(\ds \)

\(=\)

\(\ds \paren {\frac {e - \frac 1 e} {e + \frac 1 e} } i\)

multiplying denominator and numerator by $2$

\(\ds \)

\(=\)

\(\ds \paren {\frac {e^2 - 1} {e^2 + 1} } i\)

multiplying denominator and numerator by $e$

$\blacksquare$