Finding tan(90°-β)

• Aug 13th 2010, 11:48 AM
luckystar001
Finding tan(90°-β)
I have trouble doing this problem.

The problem:

Let tan β= 6
Find tan(90°-β).

• Aug 13th 2010, 12:06 PM
skeeter
Quote:

Originally Posted by luckystar001
I have trouble doing this problem.

The problem:

Let tan β= 6
Find tan(90°-β).

$\displaystyle \tan(90^\circ - \beta) = \cot{\beta} = \frac{1}{6}$
• Aug 13th 2010, 11:43 PM
Prove It
Just incase you need to show WHY $\tan{(90^{\circ} - \beta)} = \cot{\beta}$...

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

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

$= \cot{\beta}$.
• Aug 14th 2010, 03:04 AM
Quote:

Originally Posted by luckystar001
I have trouble doing this problem.

The problem:

Let tan β= 6
Find tan(90°-β).

Hi luckystar001,

Draw a right-angled triangle,

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

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

as all three angles sum to $180^o$

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

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

you can see the logic in the above posts.

A much less appetizing way is as follows...

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

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

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

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

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

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