Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1020

\[\boxed{\mathbf{1020.}}\]

\[1)\cos a \bullet \sin(6\pi - a) \bullet\]

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

\[\cos a \bullet \sin( - a) \bullet\]

\[\bullet (1 + ( - ctg\ a)^{2})) = - ctg\ a\]

\[\cos a \bullet \left( - \sin a \right) \bullet\]

\[\bullet (1 + ctg^{2}\ a)) = - ctg\ a\]

\[- \cos a \bullet \sin a \bullet\]

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

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

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

\[- ctg\ a = - ctg\ a\]

\[Что\ и\ требовалось\ доказать.\]

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

\[\bullet \frac{\sin(a - 2\pi)}{1 - \cos^{2}( - a)} = ctg\ a\]

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

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

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

\[\frac{\cos a}{\sin a} = ctg\ a\]

\[ctg\ a = ctg\ a\]

\[Что\ и\ требовалось\ доказать.\]