Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1295

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

\[Разделим\ \cos^{2}a:\]

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

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

\[= \frac{4 + 2}{1 + 6} = \frac{6}{7};\]

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

\[Разделим\cos^{2}a:\]

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

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

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

\[= \frac{2 + tg^{2}\text{\ a}}{4 + 3\ tg^{2}\text{\ a}} = \frac{2 + 2^{2}}{4 + 3 \bullet 2^{2}} =\]

\[= \frac{2 + 4}{4 + 3 \bullet 4} = \frac{6}{4 + 12} = \frac{6}{16} = \frac{3}{8}.\]