Difference of Arctangents

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Theorem

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

where $\arctan$ denotes the arctangent.


Proof

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

Then:

\((1):\quad\) \(\displaystyle \tan x\) \(=\) \(\displaystyle a\)
\((2):\quad\) \(\displaystyle \tan y\) \(=\) \(\displaystyle b\)
\(\displaystyle \map \tan {\arctan a - \arctan b}\) \(=\) \(\displaystyle \map \tan {x - y}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\tan x - \tan y} {1 + \tan x \tan y}\) Tangent of Difference
\(\displaystyle \) \(=\) \(\displaystyle \frac {a - b} {1 + a b}\) by $(1)$ and $(2)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \arctan a - \arctan b\) \(=\) \(\displaystyle \arctan \frac {a - b} {1 + a b}\)

$\blacksquare$


Also see