Question

1,已知tan（α+β）=3,tan（α-β）=5,求tan2α,tan2β的值.

Answer 1

1、

tan2a=tan(a+b+a-b)=[tan(a+b)+tan(a-b)]/(1-tan(a+b)tan(a-b))=(3+5)/(1-15)=-4/7

tan2b=tan(a+b-(a-b))=[tan(a+b)-tan(a-b)]/(1+tan(a+b)tan(a-b))=-2/(1+15)=1/8

2、

sinθ=3/5,θ∈（π/2,π）,

cosθ=-根号（1-sinθ²）=-4/5

tanθ=-3/4

tan(θ+ψ)=(tannθ+tanψ)/(1-tanθtanψ)=-2/11

tan(θ-ψ=(tanθ-tanψ)/(1+tanθtanψ)=-2