Question

求证：(sinα+cosα)^2-(sinα-cosα)^2/tanα-sinαcosα=4/tan^2α

求证：(sinα+cosα)^2-(sinα-cosα)^2/tanα-sinαcosα=4/tan^2α
算了N久都没证出来，求高手指点

Answer 1

[(sinα+cosα)² - (sinα-cosα)²] / (tanα - sinαcosα)
=(sinα + cosα + sinα - cosα)(sinα + cosα - sinα + cosα) / (sinα/cosα - sinαcosα)
=2sinα * 2cosα / [sinα (1/cosα - cosα)]
=4cosα / (1/cosα - cosα)
=4cos²α / (1-cos²α)
=4cos²α / sin²α
=4 / tan²α

Answer 2

把分子分母写清楚，用括号括起来

追问

求证：[(sinα+cosα)^2-(sinα-cosα)^2]/(tanα-sinαcosα)=4/(tan^2α)
这样么高手。。咱技术有限啊。。

追答

(sinα+cosα)^2-(sinα-cosα)^2=(1+2sinαcosα)-(1-2sinαcosα)=4sinαcosα
tanα-sinαcosα=sinα/cosα-sinαcosα=(sinα-sinαcos^2α)/cosα=sinα(1-cos^2α)/cosα=sin^3α/cosα
[(sinα+cosα)^2-(sinα-cosα)^2]/(tanα-sinαcosα)=(4sinαcosα)/(sin^3α/cosα)=4/(sin^2α/cos^2α)=4/(tan^2α)
看不懂再问我

追问

sinα(1-cos^2α)/cosα=sin^3α/cosα
↑这里是怎么算的？

追答

1-cos^2α=sin^2α