证明 2sin（П+θ）cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ）+1

题目

证明 2sin（П+θ）cosθ-1/1-2sin^2θ =tan(9П+θ)-1/tan(П+θ）+1

答案

证明：
左边=2sin（П+θ）cosθ-1/1-2sin^2θ
=(-2sinθcosθ-1)/cos2θ
=-(2sinθcosθ+sin^2 θ+cos^2 θ)/(cos^2 θ-sin^2 θ)
=-(sinθ+cosθ)^2/(cosθ-sinθ)(cosθ+sinθ)
=-(sinθ+cosθ)/(cosθ-sinθ)
=-[(sinθ/cosθ)+1]/[1-(sinθ/cosθ)]
=-(tanθ+1)/(1-tanθ)
=(tanθ+1)/(tanθ-1)
右边=tan(9П+θ)-1/tan(П+θ）+1
=(tanθ-1)/(tanθ+1)