b) operatorname(tg)(a+560^circ)-operatorname(tg)(79+a)

Para resolver essa expressão, podemos usar a identidade trigonométrica da tangente da soma de dois ângulos:\( \operatorname{tg}(A+B) = \frac{\operatorname{tg}(A) + \operatorname{tg}(B)}{1 - \operatorname{tg}(A)\operatorname{tg}(B)} \)Aplicando essa identidade à expressão dada, temos:\( \operatorname{tg}\left(a+560^{\circ}\right) = \frac{\operatorname{tg}(a) + \operatorname{tg}(560^{\circ})}{1 - \operatorname{tg}(a)\operatorname{tg}(560^{\circ})} \)E:\( \operatorname{tg}(79+a) = \frac{\operatorname{tg}(79) + \operatorname{tg}(a)}{1 - \operatorname{tg}(79)\operatorname{tg}(a)} \)Subtraindo as duas expressões, temos:\( \operatorname{tg}\left(a+560^{\circ}\right)-\operatorname{tg}(79+a) = \frac{\operatorname{tg}(a) + \operatorname{tg}(560^{\circ})}{1 - \operatorname{tg}(a)\operatorname{tg}(560^{\circ})} - \frac{\operatorname{tg}(79) + \operatorname{tg}(a)}{1 - \operatorname{tg}(79)\operatorname{tg}(a)} \)Simplificando essa expressão, chegamos à resposta correta:\( \operatorname{tg}\left(a+560^{\circ}\right)-\operatorname{tg}(79+a) = \frac{\operatorname{tg}(560^{\circ}) - \operatorname{tg}(79)}{1 - \operatorname{tg}(a)(\operatorname{tg}(560^{\circ}) - \operatorname{tg}(79))} \)