Question

Помогите решить алгебру 10 класс
$\frac{2 \sin^{2} ( \alpha ) - 1}{1 - 2 \cos^{2} ( \alpha ) } + \tan^{2} ( \alpha )$
$\frac{ \sin^{2} (x) - \cos^{2} (x) + \cos^{4} (x) }{ \cos^{2} (x) - \sin ^{2} (x) + \sin^{4} (x) }$
$\frac{ \sin^{2} (x) \cot^{2} (x) }{1 - \sin^{2} (x) } + \cot^{2} (x)$
$\frac{ \cos^{2} ( \gamma ) - \cot^{2} ( \gamma ) + 1 }{ \sin^{2} ( \gamma ) + \tan^{2} ( \gamma ) - 1 }$
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Answer 1

$1)\frac{2Sin^{2}\alpha-1}{1-2Cos^{2}\alpha}+tg^{2}\alpha=\frac{-Cos2\alpha}{-Cos2\alpha}+tg^{2}\alpha=1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha}\\\\2)\frac{Sin^{2}x-Cos^{2}x+Cos^{4}x}{Cos^{2}x-Sin^{2}x+Sin^{4}x}=\frac{Sin^{2}x-Cos^{2}x(1-Cos^{2}x)}{Cos^{2}x-Sin^{2}x(1-Sin^{2}x)}=\frac{Sin^{2}x-Cos^{2}x*Sin^{2}x }{Cos^{2}x-Sin^{2} x*Cos^{2}x}=\frac{Sin^{2}x(1-Cos^{2}x)}{Cos^{2}x(1-Sin^{2}x)}=\frac{Sin^{2}x*Sin^{2}x}{Cos^{2}x*Cos^{2}x}=\frac{Sin^{4}x }{Cos^{4}x }=tg^{4}x$

$3)\frac{Sin^{2}xCtg^{2}x}{1-Sin^{2}x }+Ctg^{2}x=\frac{Sin^{2}x*\frac{Cos^{2}x }{Sin^{2}x }}{Cos^{2}x }+Ctg^{2}x=\frac{Cos^{2}x }{Cos^{2}x }+Ctg^{2}x=1+Ctg^{2}x=\frac{1}{Sin^{2}x}$