Question

ПОМОГТТЕ ДАМ 40 БАЛОВ

Answer 1

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

$\displaystyle\bf\\3)\\\\\frac{Cos\beta }{1-Sin\beta } +\frac{1-Sin\beta }{Cos\beta } =\frac{Cos\beta \cdot Cos\beta +(1-Sin\beta )\cdot(1-Sin\beta )}{Cos\beta \cdot(1-Sin\beta )} =\\\\\\=\frac{Cos^{2} \beta +1-2Sin\beta +Sin^{2}\beta }{Cos\beta \cdot(1-Sin\beta )} =\frac{1 +1-2Sin\beta }{Cos\beta \cdot(1-Sin\beta )} =\\\\\\=\frac{2-2Sin\beta }{Cos\beta \cdot(1-Sin\beta )} =\frac{2\cdot(1-Sin\beta ) }{Cos\beta \cdot(1-Sin\beta )} =\frac{2}{Cos\beta } \\\\\\4)$
$\displaystyle\bf\\tgx+\frac{Cosx}{1+Sinx} =\frac{Sinx}{Cosx} +\frac{Cosx}{1+Sinx} =\\\\\\=\frac{Sinx\cdot(1+Sinx)+Cosx\cdot Cosx}{Cosx\cdot(1+Sinx)} =\frac{Sinx+Sin^{2}x+Cos^{2}x }{Cosx\cdot(1+Sinx)} =\\\\\\=\frac{Sinx+1}{Cosx\cdot(1+Sinx)} =\frac{1}{Cosx}$