Question

Даю 40 баллов помогите пожалуйста!!!!!!

Answer 1

$1)\frac{Sin4\alpha-Sin2\alpha}{Cos^{2} (\frac{3}{2})\alpha-Sin^{2}(\frac{3}{2})\alpha }=\frac{2Sin\frac{4\alpha-2\alpha}{2}Cos\frac{4\alpha+2\alpha}{2}}{Cos(2*\frac{3}{2})\alpha}=\frac{2Sin\alpha Cos3\alpha}{Cos3\alpha}=2Sin\alpha\\\\\\2)\frac{(Sin\alpha-Cos\alpha)^{2}}{Sin^{2}(\frac{\pi }{4}-\alpha)}=\frac{(Sin\alpha-Cos\alpha)^{2} }{(Sin\frac{\pi }{4}Cos\alpha-Cos\frac{\pi }{4}Sin\alpha)^{2}} =$

$=\frac{(Sin\alpha-Cos\alpha)^{2} }{(\frac{\sqrt{2} }{2} Cos\alpha-\frac{\sqrt{2} }{2} Sin\alpha)^{2}}=\frac{(Sin\alpha-Cos\alpha)^{2}}{(\frac{\sqrt{2} }{2})^{2}(Cos\alpha-Sin\alpha)^{2} }=\frac{1}{\frac{1}{2} }=\boxed2$

$3)\frac{Cos2\alpha+Sin^{2}\alpha}{Sin2\alpha }=\frac{Cos^{2}\alpha-Sin^{2}\alpha+Sin^{2}\alpha}{2Sin\alpha Cos\alpha}=\frac{Cos^{2}\alpha}{2Sin\alpha Cos\alpha } =\frac{Cos\alpha }{2Sin\alpha }=\boxed{\frac{1}{2}Ctg\alpha}$