\dfrac{\sin\left( \dfrac{3\pi}2+\alpha\right)}{\ctg(2\pi-\alpha)}\cdot\dfrac{\tg\left(\dfrac{\pi}2+\alpha\right)}{\sin(\pi+\alpha)}= \sin \alpha \cos \alpha \tg \alpha \ctg \alpha -\sin \alpha -\cos \alpha -\tg \alpha -\ctg \alpha :