Solve for tan(A-45°).
Recall that tan(45°) = 1
[tex]\begin{gathered} \tan (A-45\degree)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ \tan (A-45\degree)=\frac{\frac{5}{12}-1}{1+(\frac{5}{12})(1)} \\ \tan (A-45\degree)=\frac{-\frac{7}{12}}{1+\frac{5}{12}} \\ \tan (A-45\degree)=\frac{-\frac{7}{12}}{\frac{17}{12}} \\ \tan (A-45\degree)=-\frac{7}{17} \end{gathered}[/tex]
Therefore, tan(A-45°) = -7/17.
Solve for tan(B+360°)
Recall that tan(360°) = 0
[tex]\begin{gathered} \tan (B+360\degree)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \tan (B+360\degree)=\frac{\frac{5}{12}+0}{1-(\frac{5}{12})(0)} \\ \tan (B+360\degree)=\frac{\frac{5}{12}}{1-\frac{5}{12}} \\ \tan (B+360\degree)=\frac{\frac{5}{12}}{\frac{7}{12}} \\ \tan (B+360\degree)=\frac{5}{7} \end{gathered}[/tex]
Therefore, tan(B+360°) = 5/7.
