原式=√[(1-cosa)/(1+cosa)] +√[(1+cosa)/(1-cosa)]
=√[(1-cosa)²/(1-cos²a)] +√[(1+cosa)²/(1-cos²a)]
=√[(1-cosa)²/sin²a] +√[(1+cosa)²/sin²a]
=(1-cosa)/|sina|+(1+cosa)/|sina|
=2/|sina|
=-2/sina
化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a) (π
化简:根号1-cos(2π+a)/1+cos(2π+a) +根号1+cos(2π-a)/1-cos(2π-a) (π
数学人气：226 ℃时间：2019-10-23 03:22:26
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