若180°<α<360°，则化简√[(1-cosα)/(1+cosα)]-√[(1+cosα)/(1-cosα)]
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√[(1-cosα)/(1+cosα)]-√[(1+cosα)/(1-cosα)]
=√[(1-cosα)^2/(1+cosα)(1-cosα)]-√[(1+cosα)^2/(1-cosα)(1+cosα)]
=√[(1-cosα)^2/(sinα)^2]-√[(1+cosα)^2/(sinα)^2]
=-(1-cosα)/(sinα)-[-(1+cosα)/(sinα)]
=(cosα-1)/sinα+(1+cosα)/sinα
=(cosα-1+1+cosα)/sinα
=2cosα/sinα
=2cotα
=√[(1-cosα)^2/(1+cosα)(1-cosα)]-√[(1+cosα)^2/(1-cosα)(1+cosα)]
=√[(1-cosα)^2/(sinα)^2]-√[(1+cosα)^2/(sinα)^2]
=-(1-cosα)/(sinα)-[-(1+cosα)/(sinα)]
=(cosα-1)/sinα+(1+cosα)/sinα
=(cosα-1+1+cosα)/sinα
=2cosα/sinα
=2cotα
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