Question

已知tanκθ+sinθ=a,tanθ+sinθ=b,求证(a²-b²)²=16ab

Answer 1

tanθ+sinθ=a,tanθ-sinθ=b2tanθ=a+btanθ=(a+b)/2=sinθ/cosθsinθ=(a-b)/2.(1)tanθ=(a+b)/2=sinθ/cosθ=(a-b)/2/cosθcosθ=(a-b)/(a+b).(2)(1)式^2+(2)式^2:1=(a-b)^2/4+(a-b)^2/(a+b)^2=(a-b)^2(1/4+1/(a+...