I have trouble doing this problem.

The problem:

Let tan β= 6

Find tan(90°-β).

Please and thank you.

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- Aug 13th 2010, 11:48 AMluckystar001Finding tan(90°-β)
I have trouble doing this problem.

The problem:

Let tan β= 6

Find tan(90°-β).

Please and thank you. - Aug 13th 2010, 12:06 PMskeeter
- Aug 13th 2010, 11:43 PMProve It
Just incase you need to show WHY $\displaystyle \tan{(90^{\circ} - \beta)} = \cot{\beta}$...

$\displaystyle \tan{(90^{\circ} - \beta)} = \frac{\sin{(90^{\circ} - \beta)}}{\cos{(90^{\circ} - \beta)}}$

$\displaystyle = \frac{\cos{\beta}}{\sin{\beta}}$

$\displaystyle = \cot{\beta}$. - Aug 14th 2010, 03:04 AMArchie Meade
Hi luckystar001,

Draw a right-angled triangle,

label the two acute angles $\displaystyle \beta$ and $\displaystyle 90^o-\beta$

since the two acute angles must sum to $\displaystyle 90^o$

as all three angles sum to $\displaystyle 180^o$

From the definitions of $\displaystyle \displaystyle\huge\ Sin\theta=\frac{opposite}{hypotenuse}$

$\displaystyle \displaystyle\huge\ Cos\theta=\frac{adjacent}{hypotenuse}$

you can see the logic in the above posts.

A much less appetizing way is as follows...

$\displaystyle \displaystyle\huge\tan\left(90^o-\beta\right)=\frac{tan90^0+tan(-\beta)}{1-tan90^otan(-\beta)}$

As $\displaystyle \theta\rightarrow\ 90^o,\ tan\theta\rightarrow\ \infty$

hence we must take limits by dividing all terms by $\displaystyle tan90^o$

In truth we are multiplying by $\displaystyle 1=\displaystyle\huge\frac{\left(\frac{1}{tan90^o}\ right)}{\left(\frac{1}{tan90^o}\right)}$

$\displaystyle \displaystyle\huge\lim_{\theta\rightarrow90^o}\lef t(\frac{1+\frac{tan(-\beta)}{tan90^o}}{\frac{1}{tan90^o}-tan(-\beta)}\right)=\frac{1}{tan\beta}$

as $\displaystyle tan(-\beta)=-tan\beta$