求证:(1)1+tanθ/1-tanθ=tan(π/4+θ)(2)1-tanθ/1+tanθ=tan(π/4-θ)

问题描述:

求证:
(1)1+tanθ/1-tanθ=tan(π/4+θ)
(2)1-tanθ/1+tanθ=tan(π/4-θ)

1+tanθ/1-tanθ tanπ/4=1
=tanπ/4+tanθ/1-tanπ/4*tanθ
tan(π/4+θ)
同理

tan(π/4+θ)=(tanπ/4+tanθ)/(1-tanθtanπ/4) =(1+tanθ)/(1-tanθ)tan(π/4-θ)=(tanπ/4-tanθ)/(1+tanθtanπ/4) =(1-tanθ)/(1+tanθ)主要利用公式 tan(a+b)=(tana+tanb)/(1-tanata...