Задача 57755 Помогите пожалуйста....
Условие
Решение
[m]ctg α =\frac{1}{tg α }[/m]
[m]tg α =\frac{1}{3 }[/m] ⇒ [m]ctg α =\frac{1}{\frac{1}{3 }}=3[/m]
[m]tg^2 α +ctg^2 α =tg^2 α+(\frac{1}{tg α })^2=(\frac{1}{3})^2+3^2=9\frac{1}{9}[/m]
2)
[m]sin^2 α +cos^2 α =1[/m] ⇒
[m]3sin^2 α +2cos^2 α =sin^2 α +2\cdot (sin^2 α +cos^2 α)=sin^2 α+2\cdot 1[/m]
[m]sin α =\frac{1}{3}[/m]
[m]3sin^2 α +2cos^2 α =sin^2 α+2=(\frac{1}{3})^2+2=2\frac{1}{9}[/m]
2)
[m]\frac{cos β }{1+sin β }-\frac{cos β }{1-sin β }=cos β \cdot (\frac{1 }{1+sin β }-\frac{1 }{1-sin β })=cos β \cdot (\frac{1-sin β-1-sin β }{(1+sin β)(1-sin β) })=[/m]
[m]=cos β \cdot (\frac{-2sin β }{1-sin^2 β })=cos β \cdot (\frac{-2sin β }{cos^2 β })=-2tg β [/m]
4)
[m]\frac{1-4cos^2 β sin^2 β }{(cos β +sin β )^2}+2 cos β sin β=\frac{1^2-(2cos β sin β)^2 }{cos^2 β +2cos βsin β+ sin^2 β}+2 cos β sin β=\frac{(1-2cos β sin β)\cdot (1+2cos β sin β) }{1+2 cos β sin β}+2 cos β sin β=[/m]
[m]=(1-2cosβ sinβ) +2 cos β sin β=1[/m]