已知方程x^2+4ax+3a+1=0(a>1)的两根为tanα,tanβ,且α,β∈(-π/2,π/2);求tan((α+β)/2)?
tanα+tanβ=-4a
tanα*tanβ=3a+1
tan(α+β)=-4a/(1-3a-1)=4/3
tan(α+β)=2tan((α+β)/2)/((1-(tan((α+β)/2))^2)
tan((α+β)/2)=-2或tan((α+β)/2)=1/2
已知方程x^2+4ax+3a+1=0(a>1)的两根为tanα,tanβ,且α,β∈(-π/2,π/2);求tan((α+β)/2)?
tanα+tanβ=-4a
tanα*tanβ=3a+1
tan(α+β)=-4a/(1-3a-1)=4/3
tan(α+β)=2tan((α+β)/2)/((1-(tan((α+β)/2))^2)
tan((α+β)/2)=-2或tan((α+β)/2)=1/2