Question

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

Answer 1

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)