Question

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

Answer 1

a^2-b^2=(a+b)(a-b)=2tanθ*2sinθ
所以左边=16tan²θ*sin²θ
ab=tan²θ-sin²θ=sin²θ/cos²θ-sin²θ
=sin²θ(1/cos²θ-1)
=sin²θ(1-cos²θ)/cos²θ
=sin²θ*sin²θ/cos²θ
=sin²θ*tan²θ
所以16ab=16sin²θ*tan²θ
所以左边=右边

Answer 2

因为(a2-b2)=4tanθsinθ
所以(a2-b2)2=16(tanθsinθ)^2
而16ab=16(tanθ)^2-(sinθ)^2
=16(sinθ)^2/(cosθ)^2-(sinθ)^2
=16(sinθ)^2[1/(cosθ)^2-1]
=16(sinθ)^2[1-(cosθ)^2]/(cosθ)^2
=16(sinθ)^2(sinθ)^2/(cosθ)^2
=16(tanθsinθ)^2
所以(a2-b2)2=16ab