Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 467

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

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

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

\[a = \frac{\pi}{4}:\]

\[- ctg^{2}\ a = - ctg^{2}\frac{\pi}{4} = - 1^{2} = - 1.\]

\[2)\cos^{2}a + ctg^{2}\ a + \sin^{2}a =\]

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

\[\ a = \frac{\pi}{6}:\]

\[1 + ctg^{2}\ a = 1 + ctg^{2}\frac{\pi}{6} =\]

\[= 1 + \left( \sqrt{3} \right)^{2} = 1 + 3 = 4.\]

\[3)\ \frac{1}{\cos^{2}a} - 1 = \frac{1 - \cos^{2}a}{\cos^{2}a} =\]

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

\[\ a = \frac{\pi}{3}:\]

\[tg^{2}\ a = tg^{2}\frac{\pi}{3} = \left( \sqrt{3} \right)^{2} = 3.\]

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