Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 463

\[\boxed{\mathbf{463.}}\]

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

\[= \frac{\frac{1}{2} + 2}{\frac{1}{2} - 2} = \frac{0,5 + 2}{0,5 - 2} = \frac{2,5}{- 1,5} =\]

\[= - \frac{25}{15} = - \frac{5}{3}\ \]

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

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

\[= \frac{2 - 1}{2 + 1} = \frac{1}{3}\]

\[3)\ \frac{2\sin a + 3\cos a}{3\sin a - 5\cos a} =\]

\[= \frac{2 \bullet \frac{\sin a}{\cos a} + 3 \bullet \frac{\cos a}{\cos a}}{3 \bullet \frac{\sin a}{\cos a} - 5 \bullet \frac{\cos a}{\cos a}} =\]

\[= \frac{2\ tg\ a + 3}{3\ tg\ a - 5} = \frac{2 \bullet 2 + 3}{3 \bullet 2 - 5} = 7\]

\[4)\ \frac{\sin^{2}a + 2\cos^{2}a}{\sin^{2}a - \cos^{2}a} =\]

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

\[= \frac{tg^{2}\ a + 2}{tg^{2}\ a - 1} = \frac{2^{2} + 2}{2^{2} - 1} = \frac{6}{3} = 2\]