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

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- Feb 11th 2014, 05:15 PMJonroberts74arctan and arccot
why does $\displaystyle -3 cot^{-1} \frac{3}{\sqrt{x^2-9}} = 3 tan^{-1} \frac{3}{\sqrt{x^2-9}}$ ??

- Feb 11th 2014, 05:46 PMromsekRe: 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? - Feb 11th 2014, 06:01 PMJonroberts74Re: 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}} $

- Feb 11th 2014, 07:09 PMromsekRe: arctan and arccot
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, 08:28 PMJonroberts74Re: arctan and arccot
yeah I left out the constant because it's exactly that, some constant that they vary by.

- Feb 13th 2014, 09:46 AMebainesRe: 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 $\displaystyle 3 \tan^{-1}(B/A)+ 3 cot^{-1}(A/B) = \frac {3 \pi} 2$.

- Feb 13th 2014, 11:19 AMBobPRe: arctan and arccot
Aren't arctan (B/A) and arccot(A/B) the same angle ?

- Feb 13th 2014, 12:31 PMebainesRe: arctan and arccot