# Just to prove them wrong: Trigonometry

1. May 4, 2015

### mooncrater

1. The problem statement, all variables and given/known data
There is a part of solution of a question which says that:
$cot|cot^{-1}x|=cot cot^{-1}x=x$

2. Relevant equations

3. The attempt at a solution
If we put $x=-1$ in this equation then $cot^{-1}(-1)=-\pi /4$ which will have a modulus =$\pi /4$. And $cot \pi /4=1$ which is not equal to the $x$ we took. So this equation seems wrong to me. Am I correct here?

2. May 4, 2015

### Raghav Gupta

You should know the domain and ranges.
What is the range for cot-1x ?

3. May 4, 2015

### mooncrater

Hmmmm..... I think I have lost it.... it's domain is $R$ and has a range $(0, \pi)$. So $cot^{-1}(-1)=3\pi /4$ not $-\pi /4$.
I forgot its range.....
So now this equation is true for all values of $x$.
Thanks ($R\rightarrow R$ help)

4. May 5, 2015

### epenguin

If it is not equal to the x you took, it is equal to
| (the x you took) |
and there is a | | in the problem.

I think the answer you have been given is wrong and the answer is | x | .

You seem to have been given a few wrong answers and even wrong questions, or at least at the limit thereof.

5. May 5, 2015

### SammyS

Staff Emeritus
The standard range used for the arccotangent is (0, π). Thus the arccotangent function always returns a positive value.

So we have | cot-1(x) | = cot-1(x)