1.已知tanα和tan(π/4-α)是方程x^2+px+q=0的两个根，若3tanα=tan(π/4-α)求q,p

1
tanα和tan(π/4-α)是方程x^2+px+q=0的两个根,
∴tanα+tan(π/4-α)=-p/2;
tanα·tan(π/4-α)=q/2;

3tanα=tan(π/4-α),则
4tanα=-p/2;p=-8tanα
3(tanα)^2=q/2;q=6(tanα)^2.

3tanα=tan(π/4-α)
=(tanπ/4-tanα)/(1+tanπ/4·tanα)
=(1-tanα)/(1+tanα)
∴tanα=(-2±√7)/3
代入p=-8tanα,q==6(tanα)^2即可救出

2
已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值
tanα=√3(tanαtanβ+m),
tanα(1-√3tanβ)=√3m;
tanα[(√3+tanβ)/tan(π/3+β)]=√3m;
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