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,p2.已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值
问题描述：
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
2.已知tanα=√3(tanαtanβ+m),又α,β都是钝角,求α+β的值
答
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...