Question

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

Answer 1

tanθ+sinθ=a,tanθ-sinθ=b

2tanθ=a+b

tanθ=(a+b)/2=sinθ/cosθ

sinθ=(a-b)/2.1

tanθ=(a+b)/2=sinθ/cosθ=(a-b)/2/cosθ

cosθ=(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)^2

8ab=t^2-8ab

t^2=16ab

即[(a-b)(a+b)]^2=16ab

即(a^2-b^2)^2=16ab (你的题目好象少了个平方）