Страница 140 номер 465, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество: 1) \((1 - \cos \alpha)(1 + \cos \alpha) = \sin^2 \alpha\); 2) \((1 - \sin \alpha)(1 + \sin \alpha) = \cos^2 \alpha\); 3) \(\frac{\sin^2 \alpha}{1 - \sin^2 \alpha} = \operatorname{tg}^2 \alpha\); 4) \(\frac{\cos^2 \alpha}{1 - \cos^2 \alpha} = \operatorname{ctg}^2 \alpha\); 5) \(\frac{1}{1 + \operatorname{tg}^2 \alpha} + \sin^2 \alpha = 1\); 6) \(\frac{1}{1 + \operatorname{ctg}^2 \alpha} + \cos^2 \alpha = 1\).
1) \((1 - \cos \alpha)(1 + \cos \alpha) = \sin^{2} \alpha\)
\(1 - \cos^{2} \alpha = \sin^{2} \alpha;\)
2) \((1 - \sin \alpha)(1 + \sin \alpha) = \cos^{2} \alpha\)
\(1 - \sin^{2} \alpha = \cos^{2} \alpha;\)
3) \(\frac{\sin^{2} \alpha}{1 - \sin^{2} \alpha} = \operatorname{tg}^{2} \alpha\)
\(\dfrac{\sin^{2} \alpha}{\cos^{2} \alpha} = \left( \dfrac{\sin \alpha}{\cos \alpha} \right)^{2} = \operatorname{tg}^{2} \alpha;\)
4) \(\frac{\cos^{2} \alpha}{1 - \cos^{2} \alpha} = \operatorname{ctg}^{2} \alpha\)
\(\dfrac{\cos^{2} \alpha}{\sin^{2} \alpha} = \left( \dfrac{\cos \alpha}{\sin \alpha} \right)^{2} = \operatorname{ctg}^{2} \alpha;\)
5) \(\frac{1}{1 + \operatorname{tg}^{2} \alpha} + \sin^{2} \alpha = 1\)
\(\dfrac{1}{\dfrac{\cos^{2} \alpha}{\cos^{2} \alpha} + \dfrac{\sin^{2} \alpha}{\cos^{2} \alpha}} + \sin^{2} \alpha = 1,\)
\(\dfrac{\cos^{2} \alpha}{\cos^{2} \alpha + \sin^{2} \alpha} + \sin^{2} \alpha = 1,\)
\(\cos^{2} \alpha + \sin^{2} \alpha = 1;\)
6) \(\frac{1}{1 + \operatorname{ctg}^{2} \alpha} + \cos^{2} \alpha = 1\)
\(\dfrac{1}{\dfrac{\sin^{2} \alpha}{\sin^{2} \alpha} + \dfrac{\cos^{2} \alpha}{\sin^{2} \alpha}} + \cos^{2} \alpha = 1,\)
\(\dfrac{\sin^{2} \alpha}{\sin^{2} \alpha + \cos^{2} \alpha} + \cos^{2} \alpha = 1,\)
\(\sin^{2} \alpha + \cos^{2} \alpha = 1.\)
