Use the following conditions to find the exact value of tan⁡(α−β) sin⁡α=45,π2<α<π cos⁡β=513,0<β<π2

Trigonometry:
${\displaystyle \sin \alpha =\frac{4}{5},\frac{\pi }{2}<\alpha <\pi }$
${\displaystyle \cos \beta =\frac{5}{13},0<\beta <\frac{\pi }{2}}$
We need to find ${\displaystyle \tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}$
For ${\displaystyle \mathrm{\angle }\alpha }$, base ${\displaystyle =\sqrt{5^2-4^2}=3}$
for ${\displaystyle \mathrm{\angle }\beta }$, perpendicular ${\displaystyle =\sqrt{{13}^2-5^2}=12}$
Therefore ${\displaystyle \tan \alpha =-\frac{4}{3}}$
${\displaystyle \tan \beta =\frac{12}{5}}$
Therefore ${\displaystyle \tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}$
${\displaystyle =\frac{-\frac{4}{3}-\frac{12}{5}}{1+(-\frac{4}{3})\cdot \left(\frac{12}{5}\right)}}$
${\displaystyle =\frac{\frac{-56}{15}}{\frac{-33}{15}}}$
${\displaystyle =\frac{56}{33}}$