Предмет: Алгебра, автор: kimkarina953

помогите пожалуйста с алгеброй пожалуйста

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Ответы

Автор ответа: Universalka

$\displaystyle\bf\\1)\\\\\frac{1+tg\alpha }{1+Ctg\alpha } =\frac{1+tg\alpha }{1+\frac{1}{tg\alpha } } =\frac{1+tg\alpha }{\frac{tg\alpha +1}{tg\alpha } } =\frac{tg\alpha \cdot(1+tg\alpha )}{tg\alpha +1}=tg\alpha \\\\\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \Rightarrow \ \ \ tg\alpha < 0\\\\\\tg^{2} \alpha +1=\frac{1}{Cos^{2} \alpha } \\\\\\tg^{2}\alpha =\frac{1}{Cos^{2}\alpha } -1=\frac{1}{\frac{144}{169} }-1=\frac{169}{144} -1=\frac{25}{144}$

$\displaystyle\bf\\tg\alpha =-\sqrt{\frac{25}{144} }=-\frac{5}{12} \\\\\\Otvet \ : \ -\frac{5}{12} \\\\\\2)\\\\tg\alpha =\frac{5}{4} \\\\\\\frac{Sin\alpha }{Cos\alpha } =\frac{5}{4} \\\\\\4Sin\alpha =5Cos\alpha \\\\\\Sin\alpha =1,25Cos\alpha \\\\\\\frac{Sin\alpha +Cos\alpha }{Sin\alpha -Cos\alpha } =\frac{1,25Cos\alpha +Cos\alpha }{1,25Cos\alpha -Cos\alpha } =\frac{2,25Cos\alpha }{0,25Cos\alpha } =9\\\\\\Otvet \ : \ 9$

$\displaystyle\bf\\3)\\\\Cos\alpha =-\frac{1}{3} \\\\\pi < \alpha < \frac{3\pi }{2} \ \ \ \Rightarrow \ \ \ Sin\alpha < 0\\\\\\Sin\alpha =-\sqrt{1-Cos^{2}\alpha } =-\sqrt{1-\frac{1}{9} } =-\sqrt{\frac{8}{9} } =-\frac{2\sqrt{2} }{3} \\\\\\Ctg\alpha =\frac{Cos\alpha }{Sin\alpha } =-\frac{1}{3} :\Big(-\frac{2\sqrt{2} }{3} \Big)=\frac{1}{3} \cdot\frac{3}{2\sqrt{2} }=\frac{1}{2\sqrt{2} }$

$\displaystyle\bf\\\frac{Cos\alpha +Ctg\alpha }{Ctg\alpha } =\frac{-\dfrac{1}{3} +\dfrac{1}{2\sqrt{2} } }{\dfrac{1}{2\sqrt{2} } } =\frac{\dfrac{3-2\sqrt{2} }{6\sqrt{2} } }{\dfrac{1}{2\sqrt{2} } } =1-\frac{2}{3} \sqrt{2} \\\\\\Otvet \ : \ 1-\frac{2}{3} \sqrt{2} \\\\\\4)\\\\Cos^{2}\alpha -Sin^{2} \beta =0,5\\\\1-Sin^{2} \alpha -Sin^{2} \beta =0,5\\\\\\Sin^{2} \alpha -Cos^{2} \beta =Sin^{2} \alpha -(1-Sin^{2} \beta )=Sin^{2} \alpha -1+Sin^{2} \beta =$