Question

докажите,что равенство является тождеством​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{tg^{2} \alpha -Sin^{2} \alpha }{Ctg^{2} \alpha-Cos^{2}\alpha } =\frac{\dfrac{Sin^{2}\alpha }{Cos^{2}\alpha }-Sin^{2}\alpha }{\dfrac{Cos^{2} \alpha }{Sin^{2} \alpha } -Cos^{2} \alpha } =\frac{(Sin^{2}\alpha -Sin^{2} \alpha Cos^{2} \alpha)\cdot Sin^{2} \alpha}{(Cos^{2} \alpha -Cos^{2}\alpha Sin^{2}\alpha )\cdot Cos^{2} \alpha } =$

$\displaystyle\bf\\=\frac{Sin^{2} \alpha \cdot(1-Cos^{2} \alpha )\cdot Sin^{2} \alpha }{Cos^{2} \alpha \cdot(1-Sin^{2} \alpha )\cdot Cos^{2} \alpha } =\frac{Sin^{4}\alpha \cdot Sin^{2} \alpha }{Cos^{4} \alpha \cdot Cos^{2}\alpha }=\frac{Sin^{6} \alpha }{Cos^{6} \alpha } =tg^{6} \alpha \\\\\\tg^{6} \alpha =tg^{6} \alpha$

Что и требовалось доказать

$\displaystyle\bf\\2)\\\\\frac{tg\beta }{tg\beta +Ctg\beta }=\frac{\dfrac{Sin\beta }{Cos\beta } }{\dfrac{Sin\beta }{Cos\beta } +\dfrac{Cos\beta }{Sin\beta } }=\frac{\dfrac{Sin\beta }{Cos\beta } }{\dfrac{Sin^{2} \beta +Cos^{2} \beta }{Sin\beta Cos\beta } } =\frac{Sin\beta }{Cos\beta } :\frac{1}{Sin\beta Cos\beta } =\\\\\\=\frac{Sin\beta \cdot Sin\beta\cdot Cos\beta }{Cos\beta } =Sin^{2} \beta \\\\\\Sin^{2} \beta =Sin^{2} \beta$

Что и требовалось доказать

$\displaystyle\bf\\3)\\\\\frac{tg\beta }{1-tg^{2} \beta } =\frac{\dfrac{1}{Ctg\beta } }{1-\dfrac{1}{Ctg^{2} \beta } } =\frac{\dfrac{1}{Ctg\beta } }{\dfrac{Ctg^{2} \beta -1}{Ctg^{2} \beta } } =\frac{1\cdot Ctg^{2} \beta }{Ctg\beta \cdot (Ctg^{2} \beta -1)} =\\\\\\=\frac{Ctg\beta }{Ctg^{2}\beta -1 } \\\\\\\frac{Ctg\beta }{Ctg^{2}\beta -1 } =\frac{Ctg\beta }{Ctg^{2}\beta -1 }$

Что и требовалось доказать

$\displaystyle\bf\\4)\\\\\frac{Sin^{2} \alpha -Cos^{2} \alpha +Cos^{4} \alpha }{Cos^{2} \alpha -Sin^{2} \alpha +Sin^{4} \alpha } =\frac{Sin^{2} \alpha -Cos^{2} \alpha(1 -Cos^{2} \alpha )}{Cos^{2} \alpha -Sin^{2} \alpha (1-Sin^{2} \alpha) } =\\\\\\=\frac{Sin^{2} \alpha -Cos^{2} \alpha\cdot Sin^{2} \alpha }{Cos^{2} \alpha -Sin^{2} \alpha \cdot Cos^{2} \alpha } =\frac{Sin^{2} \alpha(1 -Cos^{2} \alpha) }{Cos^{2} \alpha (1-Sin^{2} \alpha) }=$

$\displaystyle\bf\\=\frac{Sin^{2}\alpha \cdot Sin^{2} \alpha }{Cos^{2} \alpha \cdot Cos^{2} \alpha } =\frac{Sin^{4}\alpha }{Cos^{4} \alpha } =tg^{4} \alpha \\\\\\tg^{4} \alpha =tg^{4} \alpha$

Что и требовалось доказать