Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1300

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\[1)\ \frac{tg^{2}\text{\ a}}{1 + ctg^{2}\text{\ a}} =\]

\[= tg^{2}\ a\ :\left( 1 + \frac{1}{tg^{2}\text{\ a}} \right) =\]

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

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

\[= tg^{4}\ a\ :\left( tg^{2}\ a + 1 \right) =\]

\[= tg^{4}\ a\ :\frac{1}{\cos^{2}a} =\]

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

\[= tg^{2}\ a \bullet \sin^{2}a;\]

\[2)\ \frac{1 + ctg^{2}\text{\ a}}{\text{ct}g^{2}\text{\ a}} =\]

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

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

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

\[3)\ \frac{tg\ \alpha - tg\ \beta}{ctg\ \alpha + ctg\ \beta} =\]

\[= \left( \frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta} \right):\left( \frac{\cos\alpha}{\sin\alpha} + \frac{\cos\beta}{\sin\beta} \right) =\]

\[= \frac{\cos{(\alpha - 2\beta) - \cos\alpha}}{\cos(\alpha + 2\beta) + \cos\alpha} =\]

\[= \frac{- 2 \bullet \sin(\alpha + \beta) \bullet \sin( - \beta)}{2 \bullet \cos(\alpha + \beta) \bullet \cos\beta} =\]

\[= tg\ (\alpha + \beta) \bullet tg\beta;\]

\[= 4\ tg\ a \bullet ctg\ a =\]

\[= 4 \bullet tg\ a \bullet \frac{1}{\text{tg\ a}} = 4 \bullet 1 = 4.\]