Sum of Arctangents/Corollary

Theorem

Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.

Then:

$\ds \map \arctan {\frac 1 n} + \map \arctan {\frac {n - 1} {n + 1} } = \frac \pi 4$

where $\arctan$ denotes the arctangent.

$\ds \arctan a + \arctan b$

$=$

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

$\ds \leadsto \ \ $

$\ds \map \arctan {\dfrac 1 n} + \map \arctan {\dfrac {n - 1} {n + 1} }$

$=$

$\ds \map \arctan {\dfrac {\dfrac 1 n + \dfrac {n - 1} {n + 1} } {1 - \paren {\dfrac 1 n} \paren {\dfrac {n - 1} {n + 1} } } }$

where $a := \dfrac 1 n$ and $b := \dfrac {n - 1} {n + 1}$

$\ds $

$=$

$\ds \map \arctan {\dfrac {\dfrac {\paren {n + 1 + n^2 - n} } {n \paren {n + 1} } } {\dfrac {\paren {n^2 + n - n + 1} } {n \paren {n + 1} } } }$

$\ds $

$=$

$\ds \map \arctan 1$

$\ds $

$=$

$\ds \frac \pi 4$

$\blacksquare$