Задача 57169 ...

exponenta

19 января 2021 г. в 17:07

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1)$(1+tg^2 α )\cdot cos^2 α =(1+\frac{sin^2 α}{cos^2 α })\cdot cos^2 α= \frac{cos^2 α+sin^2 α }{cos^2 α }\cdot cos^2 α \frac{1}{cos^2 α }\cdot cos^2 α=1$

2)$\frac{tg α +tg β }{ctg α +ctg β }=\frac{\frac {sin α }{cos α } +\frac {sin β }{cos β } }{\frac {cos α }{sin α } +\frac {cos β }{sin β } }=\frac{\frac{sin α cos β+cos α sin β }{cos α cos β }}{\frac{cos α sin β+sin αcos β }{sin αsin β }}=\frac{sin α cos β+cos α sin β }{cos α cos β }\cdot\frac{sin αsin β }{cos α sin β+sin αcos β }=\frac{sin α sin β}{ cos α cos β }=tg α tg β$


3)$\frac{cos^2 α -ctg^2 α }{sin^2 α -tg^2 α }=\frac{cos^2 α -\frac{cos^2 α }{sin^2 α } }{sin^2 α -\frac{sin^2 α}{cos^2 α} }=\frac{\frac{cos^2 α sin^2 α- cos^2 α }{sin^2 α } }{\frac{sin^2 αcos^2 α -sin^2 α}{cos^2 α} }=$

$=\frac{\frac{cos^2 α (sin^2 α- 1)}{sin^2 α } }{\frac{sin^2 α(cos^2 α -1)}{cos^2 α} }=\frac{\frac{cos^2 α (-cos^2 α)}{sin^2 α } }{\frac{sin^2 α(-sin^2 α )}{cos^2 α} }=\frac{cos^2 α (-cos^2 α)}{sin^2 α }\cdot \frac{cos^2 α}{sin^2 α(-sin^2 α )}=ctg^6 α$