в) \(\frac{\sin(-\alpha)\ctg(-\alpha)}{\cos(360^\circ-\alpha) g(180^\circ+\alpha)}\).

Используем формулы: \(\sin(-\alpha)=-\sin\alpha\), \(\ctg(-\alpha)=-\ctg\alpha\), \(\cos(360^\circ-\alpha)=\cos\alpha\), \( g(180^\circ+\alpha)= g\alpha\).

\(\frac{(-\sin\alpha)(-\ctg\alpha)}{\cos\alpha g\alpha}=\frac{\sin\alpha\cdot\frac{\cos\alpha}{\sin\alpha}}{\cos\alpha\cdot\frac{\sin\alpha}{\cos\alpha}}=\frac{\cos\alpha}{\sin\alpha}=\ctg\alpha\).

Ответ: \(\ctg\alpha\).