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)