Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1276

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\[1)\ \frac{\sin a \bullet \cos a}{\sin^{2}a - \cos^{2}a};\ tg\ a = \frac{3}{4}:\]

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

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

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

\[= - \frac{1}{2} \bullet \frac{2\ tg\ a}{1 - tg^{2}\text{\ a}} = \frac{\text{tg\ a}}{tg^{2}\ a - 1} =\]

\[= \frac{\frac{3}{4}}{\left( \frac{3}{4} \right)^{2} - 1} = \frac{\frac{3}{4}}{\frac{9}{16} - \frac{16}{16}} =\]

\[= - \frac{3}{4}\ :\frac{7}{16} = - \frac{3}{4} \bullet \frac{16}{7} =\]

\[= - \frac{3 \bullet 4}{7} = - \frac{12}{7}.\]

\[2)\sin a \bullet \cos a;\ \sin a + \cos a = \frac{1}{3}:\]

\[\sin a + \cos a = \frac{1}{3}\]

\[\left( \sin a + \cos a \right)^{2} = \left( \frac{1}{3} \right)^{2}\]

\[\sin^{2}a + \cos^{2}a + 2\sin a \bullet \cos a = \frac{1}{9}\]

\[1 + 2\sin a \bullet \cos a = \frac{1}{9}\]

\[2\sin a \bullet \cos a = - \frac{8}{9}\]

\[\sin a \bullet \cos a = - \frac{4}{9}.\]

\[Ответ:\ \ - \frac{4}{9}.\]