Question

Выражения: а) tg (180градусов-α) / ctg(90 градусов-α) б) сos²(90градусов-α)-1 / cos (180градусов -α) в) sin (π-α) / tg (π+α) г) tg (π-α)/ ctg (π/2 - α) д) sin² (180 градусов-α)-1 / cos(360градусов-α) е) ctg (π+α)-cos(90градусов+α) / tg ( 3π/2-α).

Answer 1

А) $\frac{tg(180- \alpha )}{ctg(90- \alpha )}= \frac{-tg \alpha }{tg\alpha}=-1$

б) $\frac{cos^2(90- \alpha )-1}{cos (180- \alpha )} = \frac{sin^2 \alpha -1}{cos (180- \alpha )} = \frac{sin^2\alpha -cos^2 \alpha -sin^ \alpha }{-cos \alpha}= \frac{ -cos^2 \alpha }{-cos \alpha}=cos \alpha$

в)$\frac{sin ( \pi - \alpha )}{ tg ( \pi + \alpha )} = \frac{sin \alpha }{tg \alpha } =sin \alpha * \frac{cos \alpha }{sin \alpha } =cos \alpha$

г)$\frac{tg( \pi - \alpha )}{ctg( \pi /2- \alpha) } = \frac{-tg \alpha }{tg \alpha } =-1$

д)$\frac{sin^2 ( \pi - \alpha )-1}{cos(360- \alpha )} = \frac{sin^2 \alpha -sin^2 \alpha -cos^2 \alpha }{cos \alpha } = \frac{-cos^2 \alpha }{cos \alpha } =-cos \alpha$

е) $\frac{ctg( \pi + \alpha )- cos( \pi /2+ \alpha) }{tg (3 \pi /2- \alpha )} = \frac{ctg \alpha -sin \alpha }{ctg \alpha } =1- \frac{sin^2 \alpha }{cos \alpha }$