Sum of Arccotangents
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Theorem
- $\arccot a + \arccot b = \arccot \dfrac {a b - 1} {a + b}$
where $\arccot$ denotes the arccotangent.
Proof
Let $x = \arccot a$ and $y = \arccot b$.
Then:
\(\text {(1)}: \quad\) | \(\ds \cot x\) | \(=\) | \(\ds a\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \cot y\) | \(=\) | \(\ds b\) | |||||||||||
\(\ds \map \cot {\arccot a + \arccot b}\) | \(=\) | \(\ds \map \cot {x + y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\cot x \cot y - 1} {\cot x + \cot y}\) | Cotangent of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a + b} {1 - a b}\) | by $(1)$ and $(2)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \arccot a + \arccot b\) | \(=\) | \(\ds \arccot \frac {a b - 1} {a + b}\) |
$\blacksquare$