Question

已知tanα+sinα=a，tanα-sinα=b，求证: (a²-b²)²=16ab。

Answer 1

左=(a^2-b^2)^2 =(a+b)^2·(a-b)^2 =[(tanα+sinα)+(tanα-sinα)]^2·[(tanα+sinα)-(tanα-sinα)]^2 =[2tanα]^2·[2sinα]^2 =16[tanα·sinα]^2 =16(tanα)^2·(sinα)^2; 右=16ab =16(tanα+sinα)·(tanα-sinα) =16[(tanα)^2-(sinα)^2] =16[(sinα)^2/(cosα)^2-(sinα)^2] =16(sinα)^2[(secα)^2-1] =16(sinα)^2·(tanα)^2 ∴左=右