Sum of Arctangents/Proof

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Theorem

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

Then:

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

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 Sum
\(\ds \) \(=\) \(\ds \frac {a + b} {1 - a b}\) by $(1)$ and $(2)$
\(\ds \leadsto \ \ \) \(\ds \arctan a + \arctan b\) \(=\) \(\ds \map \arctan {\frac {a + b} {1 - a b} }\)

$\blacksquare$


Sources