§ Задание 1291
\[\boxed{\mathbf{1291}\mathbf{.}}\]

\[= - \frac{2\sin a \bullet \cos a}{\cos{2a}} =\]
\[= - \frac{\sin{2a}}{\cos{2a}} = - tg\ 2a.\]
\[\boxed{\mathbf{1291}\mathbf{.}}\]

\[= - \frac{2\sin a \bullet \cos a}{\cos{2a}} =\]
\[= - \frac{\sin{2a}}{\cos{2a}} = - tg\ 2a.\]