Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1308

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\[1)\ \frac{4\sin^{2}a - \sin^{2}{2a}}{4 - 4\sin^{2}a - \sin^{2}{2a}} =\]

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

\[= \frac{4\sin^{2}a \bullet \left( 1 - \cos^{2}a \right)}{4\cos^{2}a \bullet \left( 1 - \sin^{2}a \right)} =\]

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

\[= \frac{\sin^{4}a}{\cos^{4}a} = tg^{4}\text{\ a.}\]

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

\[= \frac{3 \bullet tg^{4}\ a + 2 \bullet tg^{2}\ a - 1}{tg^{6}\ a - 2 \bullet tg^{4}\ a - 3 \bullet tg^{2}\text{\ a}} =\]

\[= \frac{3 \bullet tg^{4}\ a + 2 \bullet tg^{2}\ a - 1}{tg^{2}\ a \bullet \left( tg^{4}\ a - 2 \bullet tg^{2}\ a - 3 \right)};\]

\[3 \bullet tg^{4}\ a + 2 \bullet tg^{2}\ a - 1 = 0\]

\[D = 4 + 12 = 16\]

\[tg_{1}^{2}\ a = \frac{- 2 - 4}{2 \bullet 3} = - 1;\]

\[tg_{2}^{2}\ a = \frac{- 2 + 4}{2 \bullet 3} = \frac{2}{6} = \frac{1}{3};\]

\[3\left( tg^{2}\ a + 1 \right)\left( tg^{2} - \frac{1}{3} \right) = 0.\]

\[tg^{4}\ a - 2 \bullet tg^{2}\ a - 3 = 0\]

\[D = 4 + 12 = 16\]

\[tg_{1}^{2}\ a = \frac{2 - 4}{2} = - 1;\text{\ \ }\]

\[tg_{2}^{2}\ a = \frac{2 + 4}{2} = 3;\]

\[\left( tg^{2}\ a + 1 \right)\left( tg^{2}\ a - 3 \right) = 0.\]

\[\frac{3\left( tg^{2}\ a + 1 \right)\left( tg^{2} - \frac{1}{3} \right)}{tg^{2}\ a \bullet \left( tg^{2}\ a + 1 \right)\left( tg^{2}\ a - 3 \right)} =\]

\[= \frac{3 \bullet tg^{2} - 1}{tg^{2}\ a \bullet \left( tg^{2}\ a - 3 \right)} =\]

\[= \frac{1 - 3 \bullet tg^{2}\text{\ a}}{\left( 3 - tg^{2}\text{\ a} \right) \bullet tg^{2}\text{\ a}} =\]

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

\[= ctg\ a \bullet ctg\ 3a.\]

\[Ответ:\ \ ctg\ a \bullet ctg\ 3a.\]