Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 469

\[\boxed{\mathbf{469.}}\]

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

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

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

\[= 1 - 1 = 0\]

\[2)\ 1 - \sin^{2}a\left( 1 + ctg^{2}\text{\ a} \right) =\]

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

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

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

\[= 1 - 1 = 0\]

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

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

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

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

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

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

\[4)\ \frac{1 + tg^{2}\text{\ a}}{1 + ctg^{2}\text{\ a}} =\]

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

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

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

\[= tg^{2}\text{\ a}\]