# arctan and arccot

• Feb 11th 2014, 06:15 PM
Jonroberts74
arctan and arccot
why does $-3 cot^{-1} \frac{3}{\sqrt{x^2-9}} = 3 tan^{-1} \frac{3}{\sqrt{x^2-9}}$ ??
• Feb 11th 2014, 06:46 PM
romsek
Re: arctan and arccot
I get their difference is

\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?
• Feb 11th 2014, 07:01 PM
Jonroberts74
Re: arctan and arccot
because $\int \frac{\sqrt{x^2-9}}{x} dx = \sqrt{x^2 - 9} - 3 cot^{-1} \frac{3}{\sqrt{x^2-9}}$ and $\sqrt{x^2 - 9} + 3tan^{-1}\frac{3}{\sqrt{x^2-9}}$
• Feb 11th 2014, 08:09 PM
romsek
Re: arctan and arccot
Quote:

Originally Posted by Jonroberts74
because $\int \frac{\sqrt{x^2-9}}{x} dx = \sqrt{x^2 - 9} - 3 cot^{-1} \frac{3}{\sqrt{x^2-9}}$ and $\sqrt{x^2 - 9} + 3tan^{-1}\frac{3}{\sqrt{x^2-9}}$

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.
• Feb 11th 2014, 09:28 PM
Jonroberts74
Re: arctan and arccot
yeah I left out the constant because it's exactly that, some constant that they vary by.
• Feb 13th 2014, 10:46 AM
ebaines
Re: arctan and arccot
Given any right angle triangle with legs A and B and hypotenuse C it's pretty easy to show that: $\tan^{-1}(B/A) = \frac {\pi} 2 - cot^{-1}(A/B)$. Hence $3 \tan^{-1}(B/A)+ 3 cot^{-1}(A/B) = \frac {3 \pi} 2$.
• Feb 13th 2014, 12:19 PM
BobP
Re: arctan and arccot
Aren't arctan (B/A) and arccot(A/B) the same angle ?
• Feb 13th 2014, 01:31 PM
ebaines
Re: arctan and arccot
Quote:

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: $\tan^{-1}(B/A) = \frac {\pi} 2 - cot^{-1}(B/A)$. Hence $3 \tan^{-1}(B/A)+ 3 cot^{-1}(B/A) = \frac {3 \pi} 2$