Question

Тригонометрия, 9 класс.
Со всеми решениями, пожалуйста

Answer 1

$1)2Sin\frac{5\pi }{4}+Ctg\frac{\pi }{6}=2Sin(\pi+\frac{\pi }{4})+Ctg\frac{\pi }{6}=-2Sin\frac{\pi }{4}+Ctg\frac{\pi }{6}=-2*\frac{\sqrt{2}}{2}+\sqrt{3}=\sqrt{3}-\sqrt{2}\\\\2)\frac{1}{Ctg^{2}\alpha+1}+Cos^{2}\alpha=\frac{1}{\frac{1}{Sin^{2}\alpha}}+Cos^{2}\alpha=Sin^{2}\alpha+Cos^{2}\alpha=1$

$3)Sin(\frac{3\pi }{2}-\alpha)+Cos(\frac{\pi }{2}-\alpha)+Sin(-\alpha)+Cos(-\alpha)=-Cos\alpha+Sin\alpha-Sin\alpha+Cos\alpha=0\\\\4)\frac{1}{1-Cos\alpha}+\frac{1}{1+Cos\alpha}-2=\frac{1+Cos\alpha+1-Cos\alpha}{(1-Cos\alpha)(1+Cos\alpha)}-2=\frac{2}{1-Cos^{2}\alpha}-2=\frac{2}{Sin^{2}\alpha}-2=2*(1+Ctg^{2}\alpha)-2=2+2Ctg^{2}\alpha-2=2Ctg^{2}\alpha$

$5)\frac{3\pi }{2}<\alpha<2\pi \Rightarrow Cos\alpha >0\\\\tg\alpha=\frac{1}{Ctg\alpha}=-\frac{1}{\sqrt{3}}\\\\1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}\\\\Cos^{2}\alpha =\frac{1}{1+tg^{2}\alpha}=\frac{1}{1+(-\frac{1}{\sqrt{3}})^{2}}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}\\\\Cos\alpha=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$