1. Jul 26, 2015

### scottshannon

1. The problem statement, all variables and given/known data
Actually I have the problem solved but I do not understand how the numerator becomes equal to the denominator in the 3rd step.

2. Relevant equations
I have been looking for an identity for arctan but can't find one that seems to match.

3. The attempt at a solution
My attempt has been to derive a trig identity for arctan (A/B). I found this: arctan[(A+B)/(1âˆ’AB)]=arctanA+arctanB

2. Jul 26, 2015

### vela

Staff Emeritus
It's easier to work the other way. Use trig identities to simplify $\tan(\arctan 3x + \arctan 2x + \arctan x)$ and show it's equal to $\frac{6x(1-x^2)}{1-11x^2}$.

3. Jul 26, 2015

### scottshannon

Hmmm....ok....how do you do that?

4. Jul 26, 2015

### haruspex

Do you know a formula for expanding tan(A+B)?

5. Jul 26, 2015

### scottshannon

yes I do...I have practiced deriving it...

6. Jul 26, 2015

### scottshannon

here we are dealing with arctan though

7. Jul 26, 2015

### haruspex

So apply it to tan(arctan3x+arctan2x+arctanx)

8. Jul 26, 2015

### haruspex

Just think of arctan(whatever) as angle A.

9. Jul 26, 2015

### scottshannon

Thank you....I can expand tan (A+B) to tan (A+B+C).
What I don't understand is how did the young man who solved the problem know how to do it the way he did?

10. Jul 26, 2015

### haruspex

Please clarify: are you now saying that you can see the two are equal, but you don't understand how anyone guessed that they might be?
If the solution was created by the person who set the problem, it probably went the other way about. They started with the sum of the three arctans, in numerator and denominator, then applied the tan(A+B) expansion twice to the numerator only to disguise the equivalence.