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:<br /><br />\( \operatorname{tg}(A+B) = \frac{\operatorname{tg}(A) + \operatorname{tg}(B)}{1 - \operatorname{tg}(A)\operatorname{tg}(B)} \)<br /><br />Aplicando essa identidade à expressão dada, temos:<br /><br />\( \operatorname{tg}\left(a+560^{\circ}\right) = \frac{\operatorname{tg}(a) + \operatorname{tg}(560^{\circ})}{1 - \operatorname{tg}(a)\operatorname{tg}(560^{\circ})} \)<br /><br />E:<br /><br />\( \operatorname{tg}(79+a) = \frac{\operatorname{tg}(79) + \operatorname{tg}(a)}{1 - \operatorname{tg}(79)\operatorname{tg}(a)} \)<br /><br />Subtraindo as duas expressões, temos:<br /><br />\( \operatorname{tg}\left(a+560^{\circ}\right)-\operatorname{tg}(79+a) = \frac{\operatorname{tg}(a) + \operatorname{tg}(560^{\circ})}{1 - \operatorname{tg}(a)\operatorname{tg}(560^{\circ})} - \frac{\operatorname{tg}(79) + \operatorname{tg}(a)}{1 - \operatorname{tg}(79)\operatorname{tg}(a)} \)<br /><br />Simplificando essa expressão, chegamos à resposta correta:<br /><br />\( \operatorname{tg}\left(a+560^{\circ}\right)-\operatorname{tg}(79+a) = \frac{\operatorname{tg}(560^{\circ}) - \operatorname{tg}(79)}{1 - \operatorname{tg}(a)(\operatorname{tg}(560^{\circ}) - \operatorname{tg}(79))} \)