Atan(a/b) + atan(b/a) = pi/2

  • Thread starter Nick89
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  • #1
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Hey,

I came across this 'identity' today and was wondering if there was any algebraical explanation to this...

Basically I had to show:
[tex]\arctan( \frac{a}{b}) + \arctan( \frac{b}{a}) = \frac{\pi}{2}[/tex]
(And if necessary, (a,b) > 0 )

It is pretty easy to show when you draw a right triangle:
Code:
     /|
    /c|
   /  |
  /   |a
 /d   |
/_____|
   b
(a and b are the sides while c and d are the angles)

Now, [tex]\tan d = \frac{a}{b}[/tex] and [tex]\tan c = \frac{b}{a}[/tex] and because it is a right triangle, [tex]c + d = \frac{\pi}{2} = \arctan( \frac{a}{b}) + \arctan( \frac{b}{a})[/tex].


But I was wondering if you can also proof this algebraically?
I typed it into Maple and got [tex]\frac{1}{2} \text{signum}(\frac{a}{b}) \pi[/tex]. If my memory serves me well, signum is always 1 if a/b > 0 and -1 if a/b < 0, so if a/b > 0 this holds...

So yeah, just wondering... I can't see any way to do it algebraically...
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
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Let [itex]tan\alpha =\frac{a}{b}[/itex] and [itex]tan\beta =\frac{b}{a}[/itex]

Then consider

[tex]tan(\alpha + \beta)=\frac{tan\alpha +tan\beta}{1-tan\alpha tan\beta}[/tex]

and what that gives.
 
  • #3
555
0
Thanks, I understand now!

[tex]\tan \alpha = \frac{a}{b} \text{ , } \tan \beta = \frac{b}{a}[/tex]
[tex]\arctan \frac{a}{b} + \arctan \frac{b}{a} = \alpha + \beta[/tex]

[tex]\tan(\alpha + \beta) = \frac{ \frac{a}{b} + \frac{b}{a} }{ 1 - \frac{ab}{ab}} = \frac{ \frac{a}{b} + \frac{b}{a} }{0} = \infty[/tex]*

If [tex]\tan(\alpha + \beta) = \infty[/tex] then [tex]\alpha + \beta = \frac{\pi}{2}[/tex].

*Can you say this so easily? Shouldn't you handle the divide by zero better? I know if a denominator tends to zero, the fraction tends to infinity, but you can't divide by 0 exactly... I also can't see any way I could take a limit here?
 

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