来自陈良贵的问题
求证:(1)1+tanθ/1-tanθ=tan(π/4+θ)(2)1-tanθ/1+tanθ=tan(π/4-θ)
求证:
(1)1+tanθ/1-tanθ=tan(π/4+θ)
(2)1-tanθ/1+tanθ=tan(π/4-θ)
1回答
2020-06-30 15:43
求证:(1)1+tanθ/1-tanθ=tan(π/4+θ)(2)1-tanθ/1+tanθ=tan(π/4-θ)
求证:
(1)1+tanθ/1-tanθ=tan(π/4+θ)
(2)1-tanθ/1+tanθ=tan(π/4-θ)
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...