求证2sin(π+α)cosα-1/1-2sin²α=tan(9π+α)+1/tan(π+α)-1

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求证2sin(π+α)cosα-1/1-2sin²α=tan(9π+α)+1/tan(π+α)-1

求证2sin(π+α)cosα-1/1-2sin²α=tan(9π+α)+1/tan(π+α)-1
求证2sin(π+α)cosα-1/1-2sin²α=tan(9π+α)+1/tan(π+α)-1

求证2sin(π+α)cosα-1/1-2sin²α=tan(9π+α)+1/tan(π+α)-1
证明:左边=[-2sinαcosα-(sin²α+cos²α)]/sin²α+cos²α+-2sin²α
= -(sin²α+2sinαcosα+cos²α)/(cos²α-sin²α)
=-(sinα+cosα)²/(sinα+cosα)(cosα-sinα)
=-(sinα+cosα)/(cosα-sinα)
=(sinα+cosα)/(sinα-cosα)
=(tanα+1)/(tanα-1)
右边= (tanα+1)/(tanα-1)
所以原等式成立.