Справочник вопросов и ответов
QUOR - электронный справочник

Не знаю, как решить, весь день решаю.... посчитайте

Тег: Алгебра

1)\frac{Sin^{2}\alpha}{1-Sin^{2}\alpha}*Ctg^{2}\alpha=\frac{Sin^{2}\alpha}{Cos^{2}\alpha}*Ctg^{2}\alpha=tg^{2}\alpha*Ctg^{2}\alpha=1\\\\2)(Sin\alpha-Cos\alpha)^{2}+(Cos\alpha+Sin\alpha)^{2}=Sin^{2}\alpha-2Sin\alpha Cos\alpha+Cos^{2}\alpha+Cos^{2}\alpha+2Sin\alpha Cos\alpha+Sin^{2}\alpha=2Sin^{2}\alpha+2Cos^{2}\alpha=2(Sin^{2}\alpha+Cos^{2}\alpha)=2

3)\frac{Sin^{2}\alpha-tg^{2}\alpha}{Cos^{2}\alpha-Ctg^{2} \alpha} =\frac{Sin^{2}\alpha-\frac{Sin^{2}\alpha}{Cos^{2}\alpha}}{Cos^{2}\alpha-\frac{Cos^{2}\alpha}{Sin^{2} \alpha}}=\frac{(Sin^{2}\alpha Cos^{2}\alpha-Sin^{2}\alpha)*Sin^{2}\alpha}{(Sin^{2}\alpha Cos^{2}\alpha-Cos^{2}\alpha)*Cos^{2}\alpha} =\frac{Sin^{2}\alpha(Cos^{2}\alpha-1)*Sin^{2}\alpha}{Cos^{2}\alpha(Sin^{2}\alpha-1)*Cos^{2}\alpha}=\frac{-Sin^{6}\alpha}{-Cos^{6}\alpha}=tg^{6}\alpha

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

5)\frac{Sin^{3}\alpha-Cos^{3}\alpha}{Sin\alpha-Cos\alpha}-Sin\alpha Cos\alpha=\frac{(Sin\alpha-Cos\alpha)(Sin^{2}\alpha+Sin\alpha Cos\alpha+Cos^{2}\alpha)}{Sin\alpha-Cos\alpha}-Sin\alpha Cos\alpha=1+Sin\alpha Cos\alpha-Sin\alpha Cos\alpha=1

14