Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 494

\[\boxed{\mathbf{494.}}\]

\[1)\ tg(a + \beta)\]

\[tg\ a = - \frac{3}{4}\text{\ \ }и\ \ tg\ \beta = 2,4:\]

\[tg\ \beta = 2,4 = \frac{24}{10} = \frac{12}{5}\]

\[\text{tg}(a + \beta) = \frac{tg\ a + tg\ \beta}{1 - tg\ a \bullet tg\ \beta} =\]

\[= \frac{- \frac{3}{4} + \frac{12}{5}}{1 + \frac{3}{4} \bullet \frac{12}{5}} = \frac{- \frac{15}{20} + \frac{48}{20}}{\frac{20}{20} + \frac{36}{20}} =\]

\[= \frac{48 - 15}{20 + 36} = \frac{33}{56}\]

\[Ответ:\ \ \frac{33}{56}.\]

\[2)\ ctg(a - \beta)\ \]

\[ctg\ a = \frac{4}{3}\text{\ \ }и\ \ ctg\ \beta = - 1:\]

\[tg\ a = \frac{1}{\text{ctg\ a}} = \frac{3}{4}\text{\ \ }и\ \ \]

\[\text{tg\ }\beta = \frac{1}{\text{ctg\ }\beta} = - 1\]

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

\[= \frac{1 + tg\ a \bullet tg\ \beta}{tg\ a - tg\ \beta} = \frac{1 - \frac{3}{4}}{\frac{3}{4} + 1} =\]

\[= \frac{\frac{4}{4} - \frac{3}{4}}{\frac{3}{4} + \frac{4}{4}} = \frac{4 - 3}{3 + 4} = \frac{1}{7}\]

\[Ответ:\ \ \frac{1}{7}.\]