# Cot/Arctan

• Mar 27th 2008, 05:26 PM
TheHardcoreDave
Cot/Arctan
Math homework is not going too well...

1. cot (arctan [sq. rt.]3/x)

Any and all help is appreciated. Thanks.
• Mar 27th 2008, 05:43 PM
TheEmptySet
Attachment 5568

I hope this is clear

Good luck.
• Mar 27th 2008, 05:46 PM
teuthid
two things you need to do this:
1) $\displaystyle \cot{x}=\frac{1}{\tan{x}}$
2) $\displaystyle \tan [\tan^-1(x)]=x$

and so we have...

$\displaystyle \cot \left[ \tan^{-1} \left(\frac{\sqrt{3}}{x}\right) \right]$

$\displaystyle =\frac{1}{\tan \left[ \tan^{-1} \left(\frac{\sqrt{3}}{x}\right) \right]}$

$\displaystyle =\frac{1}{\frac{\sqrt{3}}{x}}$

$\displaystyle =\frac{x}{\sqrt{3}}=\frac{\sqrt{3}x}{3}$
• Mar 27th 2008, 09:13 PM
Soroban
Hello, TheHardcoreDave!

Quote:

$\displaystyle 1)\;\cot\left[\arctan\left(\frac{\sqrt{3}}{x}\right)\right]$

We have: . $\displaystyle \cot\underbrace{\left[\arctan\left(\frac{\sqrt{3}}{x}\right)\right]}_{\text{some angle }\theta}$

Then: .$\displaystyle \theta \:=\:\arctan\left(\frac{\sqrt{3}}{x}\right)\quad\R ightarrow\quad \tan\theta \:=\:\frac{\sqrt{3}}{x} \:=\:\frac{opp}{adj}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = \sqrt{3},\;adj = x$
Code:

                        *                     *  |                   *    |  _               *        | √3             *          |         * θ            |       * - - - - - - - - *               x

Now . . . what is $\displaystyle \cot\theta$ ?