\[\boxed{\mathbf{479.}}\]
\[- \cos a \bullet \sin a \bullet \frac{1}{\sin^{2}a} = - ctg\ a\]
\[- \frac{\cos a}{\sin a} = - ctg\ a\]
\[- ctg\ a = - ctg\ a\]
\[Что\ и\ требовалось\ доказать.\]
\[2)\ \frac{1 - \sin^{2}( - a)}{\cos(4\pi - a)} \bullet \frac{\sin(a - 2\pi)}{1 - \cos^{2}( - a)} =\]
\[= \text{ctg\ a}\]
\[\frac{1 - \left( - \sin a \right)^{2}}{\cos( - a)} \bullet \frac{\sin a}{1 - \cos^{2}a} =\]
\[= \text{ctg\ a}\]
\[\frac{1 - \sin^{2}a}{\cos a} \bullet \frac{\sin a}{\sin^{2}a} = ctg\ a\]
\[\frac{\cos^{2}a}{\cos a} \bullet \frac{1}{\sin a} = ctg\ a\]
\[\frac{\cos a}{\sin a} = ctg\ a\]
\[ctg\ a = ctg\ a\]
\[Что\ и\ требовалось\ доказать.\]