The function is given as,
[tex]\tan \alpha\text{ = }\frac{-12}{5}\text{ }(\text{ }\alpha\text{ lies in the second quadrant )}[/tex]
Calculating the value of h ,
[tex]\begin{gathered} h\text{ = }\sqrt[]{12^2+5^2} \\ h\text{ = }\sqrt[]{144\text{ + 25}} \\ h\text{ = 13} \end{gathered}[/tex]
Calculating the values of the remaining functions :
[tex]\begin{gathered} \sin \text{ }\alpha\text{ = }\frac{12}{13} \\ \cos \alpha\text{ = }\frac{-5}{13} \end{gathered}[/tex]
Now,
calculating the value of x ,
[tex]\begin{gathered} x\text{ = }\sqrt[]{5^2-3^2} \\ x\text{ = }\sqrt[]{25\text{ -9 }} \\ x\text{ = }\sqrt[]{16} \\ x\text{ = 4} \end{gathered}[/tex]
[tex]\sin \text{ }\beta=\text{ }\frac{-4}{5}[/tex]
Calculating the required functions ,
[tex]\cos \alpha\cos \beta\text{ + sin}\alpha\sin \beta\text{ = cos(}\alpha-\beta)[/tex]
Therefore,
[tex]\begin{gathered} \cos (\text{ }\alpha-\beta)\text{ =( }\frac{-5}{13})(\frac{3}{5})\text{ + (}\frac{12}{13})(\frac{-4}{5}) \\ \cos (\text{ }\alpha-\beta)\text{ = }\frac{-15}{65}\text{ }\frac{-\text{ 48}}{65} \\ \cos (\text{ }\alpha-\beta)\text{ = }\frac{-15\text{ - 48 }}{65} \\ \cos (\text{ }\alpha-\beta)\text{ = }\frac{-63}{65} \end{gathered}[/tex]
Thus the required answer is ,
[tex]\cos (\text{ }\alpha-\beta)\text{ = }\frac{-63}{65}[/tex]
