EunoR

2021-08-13

Use the following conditions to find the exact value of $\mathrm{tan}\left(\alpha -\beta \right)$
$\mathrm{sin}\alpha =\frac{4}{5},\frac{\pi }{2}<\alpha <\pi$
$\mathrm{cos}\beta =\frac{5}{13},0<\beta <\frac{\pi }{2}$

smallq9

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

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