证明:2sinα·cosα/【(sinα+cosα-1)(sinα-cosα+1)】 = (1+cosα)/sinα

2sinα·cosα/【(sinα+cosα-1)(sinα-cosα+1)】
=2sinacosa/[sin²a-(cosa-1)²]
=2sinacosa/(1-cos²a-cos²a+2cosa-1)
=2sinacosa/2cosa(1-cosa)
=sina/(1-cosa)
=sina(1+cosa)/(1-cos²a)
=sina(1+cosa)/sin²a
=(1+cosa)/sina