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

1. May 19, 2008

### Nick89

Hey,

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

$$\arctan( \frac{a}{b}) + \arctan( \frac{b}{a}) = \frac{\pi}{2}$$
(And if necessary, (a,b) > 0 )

It is pretty easy to show when you draw a right triangle:
Code (Text):
/|
/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...

2. May 19, 2008

### 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.

3. May 20, 2008

### 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?