已知tanκθ+sinθ=a,tanθ+sinθ=b,求证(a²-b²)²=16ab
问题描述：
已知tanκθ+sinθ=a,tanθ+sinθ=b,求证(a²-b²)²=16ab
答
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+b)^2)4(a+b)^2=(a-b)^2[(a+b)^2+4]令(a-b)(a+b)=t 则4(a+b)^2=t^2+4(a-b)^28ab=t^2-8abt^2=16ab即[(a-b)(a+b)]^2=16ab即(a^2-b^2)^2=16ab