I have trouble doing this problem.
The problem:
Let tan β= 6
Find tan(90°-β).
Please and thank you.
Just incase you need to show WHY $\displaystyle \tan{(90^{\circ} - \beta)} = \cot{\beta}$...
$\displaystyle \tan{(90^{\circ} - \beta)} = \frac{\sin{(90^{\circ} - \beta)}}{\cos{(90^{\circ} - \beta)}}$
$\displaystyle = \frac{\cos{\beta}}{\sin{\beta}}$
$\displaystyle = \cot{\beta}$.
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 $\displaystyle tan(-\beta)=-tan\beta$