Note the i on the first term.

This is simply e^{-5Ai}/e^{Bi }= e^{(-5A-B)i} = cos(-5A-B) + i sin(-5A-B) = cos(5A+B) - i sin(5A+B), as cos is even and sin is odd.

You could also solve this problem by turning the division into a multiplication; you will realize that the complex conjugate of cos(B)+ i sin(B), cos(B) - i sin(B), is actually its reciprocal, as cos^{2} + sin^{2} = 1, and when you start multiplying, you will recognize that the real term is the cos addition formula and the imaginary term is the sin addition formula.