Difference of Arctangents

Theorem

Let $\arctan a - \arctan b \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$

Then:

$\arctan a - \arctan b = \arctan \dfrac {a - b} {1 + a b}$

where $\arctan$ denotes the arctangent.

Proof

Let $x = \arctan a$ and $y = \arctan b$.

Then:

\(\text {(1)}: \quad\)

\(\ds \tan x\)

\(=\)

\(\ds a\)

\(\text {(2)}: \quad\)

\(\ds \tan y\)

\(=\)

\(\ds b\)

\(\ds \map \tan {\arctan a - \arctan b}\)

\(=\)

\(\ds \map \tan {x - y}\)

\(\ds \)

\(=\)

\(\ds \frac {\tan x - \tan y} {1 + \tan x \tan y}\)

Tangent of Difference

\(\ds \)

\(=\)

\(\ds \frac {a - b} {1 + a b}\)

by $(1)$ and $(2)$

\(\ds \leadsto \ \ \)

\(\ds \arctan a - \arctan b\)

\(=\)

\(\ds \arctan \frac {a - b} {1 + a b}\)

$\blacksquare$

Also see

• Sum of Arctangents

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(40)$