Question

всего 4 задания. дам 100ббал.
Срочнаа надоооо))^_^​

Answer 1

$\displaystyle\bf\\1.1\\\\\frac{Sin2\alpha }{1-Sin\Big(\dfrac{\pi }{2} -2\alpha \Big)} =\frac{Sin2\alpha }{1-Cos2\alpha } =\frac{2Sin\alpha Cos\alpha }{2Sin^{2}\alpha }=\frac{Cos\alpha }{Sin\alpha } =Ctg\alpha \\\\\\1.2\\\\\frac{Sin\Big(\dfrac{3\pi }{2}-\alpha \Big)\cdot Cos\Big(\pi +\alpha \Big)\cdot Ctg\Big(\dfrac{3\pi }{2}+\alpha \Big) }{tg\Big(2\pi -\alpha \Big)} =\\\\\\=\frac{-Cos\alpha \cdot (-Cos\alpha )\cdot(- tg\alpha )}{-tg\alpha }=Cos^{2}\alpha$

$\displaystyle\bf\\2.1\\\\\frac{Sin(-\alpha )\cdot tg(-\alpha )}{Cos(-\alpha )\cdot Ctg(-\alpha )}=\frac{-Sin\alpha \cdot(- tg\alpha) }{Cos\alpha \cdot(-Ctg\alpha )}=-tg\alpha \cdot\frac{tg\alpha }{\frac{1}{tg\alpha } } =-tg^{3}\alpha\\\\\\2.2\\\\\frac{2Sin^{2}\alpha -1 }{Sin\alpha +Cos\alpha } =\frac{1-Cos2\alpha-1 }{Sin\alpha +Cos\alpha } =-\frac{Cos2\alpha }{Sin\alpha +Cos\alpha }=$

$\displaystyle\bf\\=-\frac{Cos^{2}\alpha -Sin^{2}\alpha }{Sin\alpha +Cos\alpha }=-\frac{(Cos\alpha -Sin\alpha )\Cdot(Cos\alpha +Sin\alpha )}{Sin\alpha +Cos\alpha }=\\\\\\=-(Cos\alpha -Sin\alpha )=Sin\alpha -Cos\alpha$