ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 520

\[\boxed{\mathbf{520}\mathbf{.}}\]

\[1)\ \frac{1 - \cos{2a}}{\sin{2a}} \bullet ctg\ a = 1\]

\[\frac{2\sin^{2}a}{2\sin a \bullet \cos a} \bullet \frac{\cos a}{\sin a} = 1\]

\[1 = 1\]

\[Что\ и\ требовалось\ доказать.\]

\[2)\ \frac{\sin{2a}}{1 + \cos{2a}} = tg\ a\]

\[\frac{2\sin a \bullet \cos a}{2\cos^{2}a} = tg\ a\]

\[\frac{\sin a}{\cos a} = tg\ a\]

\[tg\ a = tg\ a\]

\[Что\ и\ требовалось\ доказать.\]

\[3)\ \frac{1 - 2\sin^{2}a}{1 + \sin{2a}} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{1 - \left( 1 - \cos{2a} \right)}{\cos^{2}a + \sin^{2}a + 2\sin a \bullet \cos a} =\]

\[= \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{\cos{2a}}{\left( \cos a + \sin a \right)^{2}} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{\left( \cos a - \sin a \right)\left( \cos a + \sin a \right)}{\left( \cos a + \sin a \right)^{2}} =\]

\[= \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{\cos a - \sin a}{\cos a + \sin a} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{\frac{\cos a}{\cos a} - \frac{\sin a}{\cos a}}{\frac{\cos a}{\cos a} + \frac{\sin a}{\cos a}} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[\frac{1 - tg\ a}{1 + tg\ a} = \frac{1 - tg\ a}{1 + tg\ a}\]

\[Что\ и\ требовалось\ доказать.\]

\[4)\ \frac{1 + \sin{2a}}{\cos{2a}} = tg\ \left( \frac{\pi}{4} + a \right)\]

\[\frac{\cos^{2}a + \sin^{2}a + 2\sin a \bullet \cos a}{\cos^{2}a - \sin^{2}a} =\]

\[= \frac{\text{tg}\frac{\pi}{4} + tg\ a}{1 - tg\frac{\pi}{4} + tg\ a}\]

\[\frac{\left( \cos a + \sin a \right)^{2}}{\left( \cos a - \sin a \right)\left( \cos a + \sin a \right)} =\]

\[= \frac{1 + tg\ a}{1 - tg\ a}\]

\[\frac{\cos a + \sin a}{\cos a - \sin a} = \frac{1 + tg\ a}{1 - tg\ a}\]

\[\frac{\frac{\cos a}{\cos a} + \frac{\sin a}{\cos a}}{\frac{\cos a}{\cos a} - \frac{\sin a}{\cos a}} = \frac{1 + tg\ a}{1 - tg\ a}\]

\[\frac{1 + tg\ a}{1 - tg\ a} = \frac{1 + tg\ a}{1 - tg\ a}\]

\[Что\ и\ требовалось\ доказать.\]