Question

помогите пожалуйста в и г​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{\sqrt{2} Cos\alpha-2Cos\Big(\dfrac{\pi }{4} +\alpha \Big) }{2Sin\Big(\dfrac{\pi }{4} +\alpha \Big)-\sqrt{2}Sin\alpha } =\frac{\sqrt{2} \Big[Cos\alpha-\sqrt{2} Cos\Big(\dfrac{\pi }{4} +\alpha \Big)\Big] }{\sqrt{2}\Big[\sqrt{2} Sin\Big(\dfrac{\pi }{4} +\alpha \Big)-Sin\alpha\Big] } =$

$\displaystyle\bf\\=\frac{Cos\alpha -\sqrt{2}\cdot\Big( Cos\dfrac{\pi }{4}\cdot Cos\alpha -Sin\dfrac{\pi }{4} \Cdot Sin\alpha \Big)}{\sqrt{2} \cdot\Big(Sin\dfrac{\pi }{4}\cdot Cos\alpha+Sin\alpha \cdot Cos\dfrac{\pi }{4} \Big)-Sin\alpha } =\\\\\\=\frac{Cos\alpha -\sqrt{2}\cdot\Big( \dfrac{\sqrt{2} }{2} \cdot Cos\alpha -\dfrac{\sqrt{2} }{2} \cdot Sin\alpha \Big)}{\sqrt{2} \cdot\Big(\dfrac{\sqrt{2} }{2} \cdot Cos\alpha+Sin\alpha \cdot \dfrac{\sqrt{2} }{2} \Big)-Sin\alpha } =$

$\displaystyle\bf\\=\frac{Cos\alpha -Cos\alpha +Sin\alpha }{Cos\alpha +Sin\alpha -Sin\alpha } =\frac{Sin\alpha }{Cos\alpha } =tg\alpha$

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