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Ответ:

danilpasynkov1

03.03.2021 16:10

$5)\frac{1-Ctg\alpha }{1-tg\alpha }=\frac{1-\frac{1}{tg\alpha}}{1-tg\alpha }=\frac{tg\alpha-1 }{tg\alpha(1-tg\alpha)} =-\frac{1}{tg\alpha }=\boxed{-Ctg\alpha}\\\\\\6)\frac{tg\alpha -1}{Ctg\alpha-1 }=\frac{tg\alpha-1 }{\frac{1}{tg\alpha}-1 }=\frac{tg\alpha(tg\alpha-1)}{1-tg\alpha }=\boxed{-tg\alpha}\\\\\\7)\frac{1}{1+Sin\alpha } -\frac{1}{1-Sin\alpha }=\frac{1-Sin\alpha-1-Sin\alpha}{(1+Sin\alpha)(1-Sin\alpha)}=-\frac{2Sin\alpha}{1-Sin^{2}\alpha}=\boxed{-\frac{2Sin\alpha }{Cos^{2}\alpha}}$

$8)\frac{Ctg\alpha+1 }{tg\alpha+1 }=\frac{\frac{1}{tg\alpha}+1}{tg\alpha+1 }=\frac{1+tg\alpha }{tg\alpha(tg\alpha+1)}=\frac{1}{tg\alpha }=\boxed{Ctg\alpha}\\\\\\9)\frac{Sin\alpha }{1+Cos\alpha}+Ctg\alpha=\frac{Sin\alpha }{1+Cos\alpha}+\frac{Cos\alpha }{Sin\alpha }=\frac{Sin^{2}\alpha+Cos\alpha+Cos^{2}\alpha}{Sin\alpha(1+Cos\alpha)}=\\\\=\frac{1+Cos\alpha }{Sin\alpha(1+Cos\alpha)}=\boxed{\frac{1}{Sin\alpha}}$

$10)\frac{Cos\alpha }{1-Sin\alpha }-tg\alpha =\frac{Cos\alpha }{1-Sin\alpha }-\frac{Sin\alpha }{Cos\alpha } =\frac{Cos^{2}\alpha-Sin\alpha+Sin^{2}\alpha}{Cos\alpha(1-Sin\alpha)}=\\\\=\frac{1-Sin\alpha }{Cos\alpha(1-Sin\alpha)}=\boxed{\frac{1}{Cos\alpha}} \\\\\\11)\frac{Sin\beta }{1-Cos\beta}+\frac{Sin\beta }{1+Cos\beta}=\frac{Sin\beta+Sin\beta Cos\beta+Sin\beta-Sin\beta Cos\beta}{(1-Cos\beta)(1+Cos\beta) }=\frac{2Sin\beta }{1-Cos^{2}\beta} =\\\\=\frac{2Sin\beta }{Sin^{2}\beta}$

$=\boxed{\frac{2}{Sin\beta }}\\\\\\12)\frac{Cos\beta }{1+Sin\beta}+\frac{Cos\beta }{1-Sin\beta }=\frac{Cos\beta-Sin\beta Cos\beta+Cos\beta+Sin\beta Cos\beta}{(1+Sin\beta)(1-Sin\beta)}=\\\\=\frac{2Cos\beta }{1-Sin^{2}\beta}=\frac{2Cos\beta }{Cos^{2}\beta}=\boxed{\frac{2}{Cos\beta}}$