calculating $\tan^{-1}(a)+\tan^{-1}(b)$

Asked Nov 23 '17 at 18:46

Active Nov 23 '17 at 19:37

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I am trying to calculate the sum of the angles between the position vectors of two imaginary numbers and the real axis without using the $\operatorname{arg} function.

The numbers are:

$4+i$ and $5+3i$

According to Wolfarm the angles $\tan^{-1}(\frac14)+\tan^{-1}(\frac35)$ are equal to $\tan^{-1}(1)$

My teacher told me that there is a way to multiply vectors in $2D$ space to get the sum of their angle as an expression of $\arctan$ ...

edited Nov 23 '17 at 19:37

amWhy

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asked Nov 23 '17 at 18:46

Lendion

3

$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$, so $\tan(\tan^{-1}a+\tan^{-1}b)=\frac{a+b}{1-ab}$. – Angina Seng Nov 23 '17 at 18:48
https://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125 – lab bhattacharjee Nov 23 '17 at 18:54
Note that $\arg(a+bi)=\textrm{atan2}(b,a)=\tan^{-1}(b/a)$. – Somos Nov 23 '17 at 18:55
See this page – Bernard Nov 23 '17 at 18:56
see here https://www.quora.com/What-is-the-sum-of-two-inverse-tangents – Dr. Sonnhard Graubner Nov 23 '17 at 19:00
@Lord Shark the Unknown that is incorrect it should be arctan that but i do not know why – Lendion Nov 23 '17 at 19:02

calculating $\tan^{-1}(a)+\tan^{-1}(b)$

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