已知tanαtanβ=根号3/3,求证(2-cos2α)(2-cos2β)=3

方法一：由tanαtanβ＝√3/3得：√3sinαsinβ＝cosαcosβ
两边平方得：3sin²αsin²β＝cos²αcos²β
∴3(1-cos2α)(1-cos2β)＝(1+cos2α)(1+cos2β)
整理得：1-2(cos2α+cos2β+cos2αcos2β)＝0
∴2(cos2α+cos2β+cos2αcos2β)=1
(2-cos2α)(2-cos2β)＝4-2(cos2α+cos2β+cos2αcos2β)
＝4-1
＝3
方法二：(2-cos2α)(2-cos2β)＝[2-(1-tan2α)(1+tan2α)] [2-(1-tan2β)(1+tan2β)]
＝[(1+3tan2α)(1+tan2α)] [(1+3tan2β)(1+tan2β)]
＝(1+3tan2α+3tan2β+9 tan2αtan2β)/(1+tan2α+tan2β+ tan2αtan2β)
＝(4+3tan2α+3tan2β)/(4/3+tan2α+tan2β)＝3