# Trigonometric proof using derivatives

1. Jan 20, 2012

### Pole

1. The problem statement, all variables and given/known data
Prove that (for every x smaller than -1)
$\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}}+arccotx=\pi$

2. Relevant equations

3. The attempt at a solution
So i split the formula into two parts:
$\displaystyle \frac{1}{2}arctan\frac{2x}{1-x^{2}}$ and $\displaystyle \pi-arccotx$
Calculated their derivatives separately
And they are both equal to $\displaystyle \frac{1}{x^{2}+1}$

And I'm wondering whether it's the end of the task or should I prove some other things, especially something connected to the main assumption of x<-1 or any other.

2. Jan 20, 2012

### micromass

Staff Emeritus
If $f^\prime=g^\prime$, then this doesn't mean f=g necessarily. But it does mean that f=g+C with C a real constant/

Note that the above only holds if the domain of f and g is an interval.

3. Jan 21, 2012

### Pole

Thank You, now I know that it isn't sufficient.
But I can't really come up with a complete solution.
What should I add ? So that the task would be marked well
How to calculate the value of that constant?