回復 1# Exponential 的帖子
\( \cos{50^{\circ}} = - \cos{130^{\circ}} = - \cos{(60^{\circ}+70^{\circ})} = - \cos{60^{\circ}} \cos{70^{\circ}} + \sin{60^{\circ}} \sin{70^{\circ}} \)
所求：\( \displaystyle \frac{ \cos{50^{\circ}} }{ \sin{60^{\circ}} \sin{70^{\circ}} } + \frac{ \cos{60^{\circ}} }{ \sin{50^{\circ}} \sin{70^{\circ}} } + \frac{ \cos{70^{\circ}} }{ \sin{50^{\circ}} \sin{60^{\circ}} } \)
\( = \displaystyle \frac{ - \cos{60^{\circ}} \cos{70^{\circ}} + \sin{60^{\circ}} \sin{70^{\circ}} }{ \sin{60^{\circ}} \sin{70^{\circ}} } + \frac{ - \cos{50^{\circ}} \cos{70^{\circ}} + \sin{50^{\circ}} \sin{70^{\circ}} }{ \sin{50^{\circ}} \sin{70^{\circ}} } + \frac{ - \cos{50^{\circ}} \cos{60^{\circ}} + \sin{50^{\circ}} \sin{60^{\circ}} }{ \sin{50^{\circ}} \sin{60^{\circ}} } \)
\( = \displaystyle - \cot{60^{\circ}} \cot{70^{\circ}} + 1 - \cot{50^{\circ}} \cot{70^{\circ}} + 1 - \cot{50^{\circ}} \cot{60^{\circ}} + 1 = 3 - 1 = 2 \)
