Question

Алгебра. Задание 266.

Answer 1

Ответ:

$1)cos \alpha \times tg \alpha - 2sin \alpha = \\ = cos \alpha \times \frac{sin \alpha }{cos \alpha } - 2sin \alpha = \\ = sin \alpha - 2sin \alpha = - sin \alpha$

$2)cos \alpha - sin \alpha \times ctg \alpha = \\ = cos \alpha - sin \alpha \times \frac{ \cos \alpha }{sin \alpha } = \\ = cos \alpha - cos \alpha = 0$

$3) \frac{ {sin}^{2} \alpha }{1 + cos \alpha } = \frac{1 - {cos}^{2} \alpha }{1 + cos \alpha } = \\ = \frac{(1 - cos \alpha )(1 + cos \alpha )}{1 + cos \alpha } = \\ = 1 - cos \alpha$

$4) \frac{ {cos}^{2} \alpha }{1 - \sin \alpha } = \frac{1 - {sin}^{2} \alpha }{1 - sin \alpha } = \\ = \frac{(1 + sin \alpha )(1 - sin \alpha )}{1 - sin \alpha } = \\ = 1 + sin \alpha$

$5) \frac{tg \alpha \times cos \alpha }{ {sin}^{2} \alpha } = \frac{ \frac{sin \alpha }{cos \alpha } \times cos \alpha }{ {sin}^{2} \alpha } = \\ = \frac{sin \alpha }{ {sin}^{2} \alpha } = \frac{1}{sin \alpha }$