§ Задание 1305
\[\boxed{\mathbf{1305}\mathbf{.}}\]

\[= - \sin^{2}x - tg\ x \bullet ( - ctg\ x) =\]
\[= - \sin^{2}x + tg\ x \bullet \frac{1}{\text{tg\ x}} =\]
\[= 1 - \sin^{2}x = \cos^{2}x.\]
\[\boxed{\mathbf{1305}\mathbf{.}}\]

\[= - \sin^{2}x - tg\ x \bullet ( - ctg\ x) =\]
\[= - \sin^{2}x + tg\ x \bullet \frac{1}{\text{tg\ x}} =\]
\[= 1 - \sin^{2}x = \cos^{2}x.\]