\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
Задание

Выбери верный ответ

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

  • \(\sin \alpha\)
  • \(\cos \alpha\)
  • \(\tg \alpha\)
  • \(\ctg \alpha\)
  • \(-\sin \alpha\)
  • \(-\cos \alpha\)
  • \(-\tg \alpha\)
  • \(-\ctg \alpha\)