1+2sin(π-α)cos(4π-α)/sin²(π+α)-cos²(π-α)=tanα+1/ta

证明：
左边=[1+2sin(π-α)cos(4π-α)]/[sin²(π+α)-cos²(π-α)]
=(1+2sinαcosα)/(sin²α-cos²α)
=(sin²α+cos²α+2sinαcosα)/(sin²α-cos²α)
=(sinα+cosα)²/[(sinα+cosα)(sinα-cosα)]
=(sinα+cosα)/(sinα-cosα) 分子、分母同除以cosα
=(tanα+1)/(tanα-1)
=右边
∴[1+2sin(π-α)cos(4π-α)]/[sin²(π+α)-cos²(π-α)]=(tanα+1)/(tanα-1)

左式=（1+2sinαcosα)/[sin²(-α)-(-cosα)²]=(sin²α+2sinαcosα+cos²α)/(sin²α-cos²α)
=(sinα+cosα)²/(sin²α-cos²α)=(sinα+cosα)/(sinα-cosα)=(sinα/cosα+1)/(sinα/cosα-1)
=(tanα+1)/(tanα-1)=右式