# Thread: arctan and arccot

1. ## arctan and arccot

why does $\displaystyle -3 cot^{-1} \frac{3}{\sqrt{x^2-9}} = 3 tan^{-1} \frac{3}{\sqrt{x^2-9}}$ ??

2. ## Re: arctan and arccot

I get their difference is

\displaystyle \begin{align*}&\frac{3\pi}{2} \;\; \{x: \left|x\right| < 3 \} \\ \\ &\frac{-3\pi}{2} \;\; \{x: \left|x \right| \geq 3 \} \end{align}

so I wouldn't call them equal. Why do you think they are?

3. ## Re: arctan and arccot

because $\displaystyle \int \frac{\sqrt{x^2-9}}{x} dx = \sqrt{x^2 - 9} - 3 cot^{-1} \frac{3}{\sqrt{x^2-9}}$ and $\displaystyle \sqrt{x^2 - 9} + 3tan^{-1}\frac{3}{\sqrt{x^2-9}}$

4. ## Re: arctan and arccot

Originally Posted by Jonroberts74
because $\displaystyle \int \frac{\sqrt{x^2-9}}{x} dx = \sqrt{x^2 - 9} - 3 cot^{-1} \frac{3}{\sqrt{x^2-9}}$ and $\displaystyle \sqrt{x^2 - 9} + 3tan^{-1}\frac{3}{\sqrt{x^2-9}}$
Ok, I show that the derivatives of each of these expressions do indeed equal the integrand but remember that there is a constant of integration when you integrate. So the two expressions you have that integral equal to can in fact differ by a constant, which they apparently do.

5. ## Re: arctan and arccot

yeah I left out the constant because it's exactly that, some constant that they vary by.

6. ## Re: arctan and arccot

Given any right angle triangle with legs A and B and hypotenuse C it's pretty easy to show that: $\displaystyle \tan^{-1}(B/A) = \frac {\pi} 2 - cot^{-1}(A/B)$. Hence $\displaystyle 3 \tan^{-1}(B/A)+ 3 cot^{-1}(A/B) = \frac {3 \pi} 2$.

7. ## Re: arctan and arccot

Aren't arctan (B/A) and arccot(A/B) the same angle ?

8. ## Re: arctan and arccot

Originally Posted by BobP
Aren't arctan (B/A) and arccot(A/B) the same angle ?
My bad, thanks for catching the error. What I should have wriiten is:

it's pretty easy to show that: $\displaystyle \tan^{-1}(B/A) = \frac {\pi} 2 - cot^{-1}(B/A)$. Hence $\displaystyle 3 \tan^{-1}(B/A)+ 3 cot^{-1}(B/A) = \frac {3 \pi} 2$