Вопрос:

Ответ:

\[= \frac{\sin(59{^\circ} + 61{^\circ})}{\cos(58{^\circ} + 62{^\circ})} = \frac{\sin{120{^\circ}}}{\cos{120{^\circ}}} =\]

\[= tg\ 120{^\circ} < 0\]

\[B = \frac{\sin{36{^\circ}} + \cos{36{^\circ}}}{\cos{18{^\circ}} - \sin{18{^\circ}}} > 0;\ \ \]

\[так\ как\ \sin{36{^\circ}}\ и\ \]

\[\cos{36{^\circ}} > 0 \Longrightarrow\]

\[\Longrightarrow \sin{36{^\circ}} + \cos{36{^\circ}} > 0\ \ \]

\[для\ 0 < a < 45{^\circ} \Longrightarrow\]

\[\Longrightarrow \cos a > \sin a \Longrightarrow\]

\[\Longrightarrow \cos{18{^\circ}} - \sin{18{^\circ}} > 0.\ \]

\[A < B.\]