- #1

- 555

- 0

## Main Question or Discussion Point

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:

(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...

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
```

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...