Question

Помогите пожалуйста по алгебре

Answer 1

$1)\frac{Sin\alpha+Cos\alpha}{Sin\alpha-Cos\alpha}=\frac{\frac{Sin\alpha }{Cos\alpha}+\frac{Cos\alpha }{Cos\alpha}}{\frac{Sin\alpha }{Cos\alpha}-\frac{Cos\alpha }{Cos\alpha}}=\frac{tg\alpha+1 }{tg\alpha-1}=\frac{\frac{5}{4}+1 }{\frac{5}{4}-1 }=\frac{9}{4}:\frac{1}{4} =\frac{9}{4}*4=\boxed9\\\\\\2)\frac{Cos\alpha+Ctg\alpha}{Ctg\alpha }=\frac{Cos\alpha }{Ctg\alpha }+\frac{Ctg\alpha }{Ctg\alpha }=\frac{Cos\alpha }{\frac{Cos\alpha }{Sin\alpha}}+1=Sin\alpha +1$

$\pi< \alpha<\frac{3\pi }{2} \ \Rightarrow \ Sin\alpha<0\\\\Sin\alpha=-\sqrt{1-Cos^{2}\alpha}=-\sqrt{1-(-\frac{1}{3})^{2}}=-\sqrt{1-\frac{1}{9}}=-\sqrt{\frac{8}{9} }=-\frac{2\sqrt{2} }{3} \\\\Sin\alpha+1=-\frac{2\sqrt{2} }{3}+1=\boxed{\frac{3-2\sqrt{2}}{3}}\\\\\\3)Sin^{2}\alpha-Cos^{2}\beta=1-Cos^{2}\alpha -(1-Sin^{2}\beta)=1-Cos^{2}\alpha-1+Sin^{2}\beta=\\\\=Sin^{2}\beta-Cos^{2}\alpha=-( Cos^{2}\alpha-Sin^{2}\beta)=\boxed{-0,5}$