Question

Решите пожалуйста эти задание

Answer 1

$\displaystyle\bf\\7)\\\\\frac{Cos2\alpha +1}{Cos^{2}\alpha } =\frac{2Cos^2\alpha -1+1}{Cos^{2}\alpha } =\frac{2Cos^{2} \alpha }{Cos^{2} \alpha }=2\\\\\\8)\\\\\pi < \alpha < \frac{3\pi }{2} \ \ \ \Rightarrow \ \ \ Cos\alpha < 0\\\\\\Cos\alpha =-\sqrt{1-Sin^{2}\alpha }=-\sqrt{1-\Big(-\frac{3}{5} \Big)^{2} } =-\sqrt{1-\frac{9}{25} } =\\\\\\=-\sqrt{\frac{16}{25} } =-\frac{4}{5} \\\\\\Cos2\alpha =1-2Sin^{2}\alpha =1-2\cdot \Big(-\frac{3}{5}\Big)^{2}=1-2\cdot\frac{9}{25} =$

$\displaystyle\bf\\=1-\frac{18}{25} =\frac{7}{25} =0,28\\\\\\\boxed{Cos2\alpha =0,28}\\\\\\tg\alpha =\frac{Sin\alpha }{Cos\alpha } =-\frac{3}{5} :\Big(-\frac{4}{5} \Big)=\frac{3}{5}\cdot\frac{5}{4} =\frac{3}{4} =0,75\\\\\\\boxed{tg\alpha =0,75}\\\\\\9)\\\\\frac{2Cos^{2} \alpha \cdot tg\alpha }{Sin^{2} \alpha -Cos^{2} \alpha } =\frac{2Cos^{2} \alpha \cdot \dfrac{Sin\alpha }{Cos\alpha } }{-Cos2\alpha } =\frac{2Sin\alpha Cos\alpha }{-Cos2\alpha } =$

$\displaystyle\bf\\=\frac{Sin2\alpha }{-Cos2\alpha } =-tg\alpha \\\\\\-tg\alpha =-tg\alpha$

Тождество доказано