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Answer 1

`tan(\alpha - \beta) = \cfrac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)} \\\\[-0.35em] ~\dotfill\\\\ cos(\alpha )=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the opposite} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies √(c^2 - a^2)=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`

`\pm√(5^2 - 3^2)=b\implies \pm 4=b\implies \stackrel{IV~Quadrant}{-4=b}~\hfill tan(\alpha)=\cfrac{\stackrel{opposite}{-4}}{\underset{adjacent}{3}} \\\\[-0.35em] ~\dotfill`

`tan(\alpha - \beta) \implies \cfrac{\stackrel{tan(\alpha)}{-(4)/(3)}- \stackrel{tan(\beta)}{(4)/(3)}}{1+ \stackrel{tan(\alpha)}{\left( -(4)/(3) \right)}\stackrel{tan(\beta)}{\left( (4)/(3)\right)}}\implies \cfrac{~~ (-8 )/(3 ) ~~}{1-(16)/(9)}\implies \cfrac{~~ (-8 )/(3 ) ~~}{(-7)/(9)} \\\\\\ \cfrac{-8}{3}\cdot \cfrac{9}{-7}\implies \cfrac{24}{7}\implies {\Large \begin{array}{llll} 3(3)/(7) \end{array}}`