Is there an algebraic explanation for the identity atan(a/b) + atan(b/a) = pi/2?

[Nick89]

Hey,

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

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

It is pretty easy to show when you draw a right triangle:

     /|
    /c|
   /  |
  /   |a
 /d   |
/_____|
   b

(a and b are the sides while c and d are the angles)

Now, $$\tan d = \frac{a}{b}$$ and $$\tan c = \frac{b}{a}$$ and because it is a right triangle, $$c + d = \frac{\pi}{2} = \arctan( \frac{a}{b}) + \arctan( \frac{b}{a})$$.But I was wondering if you can also proof this algebraically?
I typed it into Maple and got $$\frac{1}{2} \text{signum}(\frac{a}{b}) \pi$$. 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...

 

[rock.freak667]

Let $tan\alpha =\frac{a}{b}$ and $tan\beta =\frac{b}{a}$

Then consider

$$tan(\alpha + \beta)=\frac{tan\alpha +tan\beta}{1-tan\alpha tan\beta}$$

and what that gives.

 

[Nick89]

Thanks, I understand now!

$$\tan \alpha = \frac{a}{b} \text{ , } \tan \beta = \frac{b}{a}$$
$$\arctan \frac{a}{b} + \arctan \frac{b}{a} = \alpha + \beta$$

$$\tan(\alpha + \beta) = \frac{ \frac{a}{b} + \frac{b}{a} }{ 1 - \frac{ab}{ab}} = \frac{ \frac{a}{b} + \frac{b}{a} }{0} = \infty$$*

If $$\tan(\alpha + \beta) = \infty$$ then $$\alpha + \beta = \frac{\pi}{2}$$.

*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?