Cot/Arctan

**TheHardcoreDave** · March 27th 2008, 05:26 PM

Math homework is not going too well...

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

Any and all help is appreciated. Thanks.

**TheEmptySet** · March 27th 2008, 05:43 PM

I hope this is clear

Good luck.

**teuthid** · March 27th 2008, 05:46 PM

two things you need to do this:
1) $\cot{x}=\frac{1}{\tan{x}}$
2) $\tan [\tan^-1(x)]=x$

and so we have...

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

$<br /> =\frac{1}{\tan \left[ \tan^{-1} \left(\frac{\sqrt{3}}{x}\right) \right]}$

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

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

**Soroban** · March 27th 2008, 09:13 PM

Hello, TheHardcoreDave!

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

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

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

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

Code:

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

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

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