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atan(x)+atan(y)=atan((x+y)/(1-xy))

x,y=?

 Aug 7, 2016

Best Answer 

 #1
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+4

Hi

x>0 , y>0 and x*y<1

 Aug 7, 2016
 #1
avatar
+4
Best Answer

Hi

x>0 , y>0 and x*y<1

Guest Aug 7, 2016
 #2
avatar+118658 
+5

     

atan(x)+atan(y)=atan((x+y)/(1-xy))

x,y=?

 

\(atan(x)+atany=atan(\frac{x+y}{1-xy})\\ let\;\;x=tanp\;\;\;and\;\;y=tanq\\ atan(tanp)+atan(tanq)=atan(\frac{tanp+tanq}{1-tanptanq})\\ p+q=atan(tan(p+q))\\ p+q=p+q\\ \)

 

There are restrictions on the domain and range but what I have discovered is that, allowing for the restrictions, this is an identity.   (it is always true)

Determining the restricions is more confusing.

 

An obvious restricion is that 1-xy cannot be 0

\(xy\ne1\)

 

I can't get my head around the rest of the restrictions but here is what Wolfram alpha has to say about it..

 

https://www.wolframalpha.com/input/?i=atan(x)%2Batan(y)%3Datan((x%2By)%2F(1-xy))

 Aug 8, 2016
 #3
avatar+9665 
0

What Melody explained means x and y could be anything unless x times y = 1.

Example x = .5 and y = 2 is impossible because the right hand side will be undefined. That will become atan of an undefined number which makes it more complicated. Actually I think this thing I already complicated enough XD

MaxWong  Aug 8, 2016

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