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）