题目

已知tan(π/12+α)=根号2,tan(β-π/3)=2根号2,求tan（α+β）的值

提问时间：2021-03-02

答案

tan(α+β-π/4)
=tan(π/12+α+β-π/3)
=[tan(π/12+α)+tan(β-π/3)]/[1-tan(π/12+α)tan(β-π/3)]
=(√2+2√2)/[1-√2*2√2]
=3√2/(1-4)
=-√2
tan(α+β-π/4)
=[tan(α+β)-tanπ/4]/[1+tan(α+β)tanπ/4]
=[tan(α+β)-1]/[1+tan(α+β)]
[tan(α+β)-1]/[1+tan(α+β)]=-√2
tan(α+β)-1=-√2[1+tan(α+β)]
tan(α+β)-1=-√2-√2tan(α+β)
tan(α+β)+√2tan(α+β)=1-√2
(1+√2)tan(α+β)=1-√2
tan(α+β)=(1-√2)/(1+√2)
tan(α+β)=(1-√2)^2/(1+√2)(1-√2)
tan(α+β)=-(1-√2)^2
tan(α+β)=2√2-3