求证 sin^6α+cos^6α=(tan^2α+1-3sin^2α)/(tan^2α+1)
问题描述：
求证 sin^6α+cos^6α=(tan^2α+1-3sin^2α)/(tan^2α+1)
2.已知π/2＜α＜π，若cosα-sinα=-根号5/5，求（sinα+cosα+2sinαcosα）/（1-tanα）的值
答
左边= sin^6α+cos^6α=( sin^2α+cos^2α)( sin^4α- sin^2αcos^2α+cos^4α)= sin^4α- sin^2αcos^2α+cos^4α= sin^4α+2sin^2αcos^2α+cos^4α-3sin^2αcos^2α=(sin^2α+cos^2α)²-3sin^2αcos^2α=1--...能不能帮我再看一下另一题，感激不尽~