[cos2α/(1+sinα2)]*[(1+tanα)/(1-tanα)]=?

sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β=（1-cos2α）/2*（1-cos2β）/2+（1+cos2α）/2*（1+cos2β）/2-1/2cos2α*cos2β=1/4（1+cos2α*cos2β-cos2α-cos2β）+1/4（1+cos2α*cos2β+cos2α+cos2β）-1/2cos2α*cos2β=1/4+1/ 证明:1-cos2α/sinα=2sinα ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 急 ((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 化简(sinα-cosα)^2-1/-cos2α sinα-cosα=1/2 求cos2α 2cos2α-1/1-2sinα2 化简 化简(sinα)^2*(sinβ)^2+(cosα)^2(cosβ)^2-1/2cos2αcos2β 化简:sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β 化解sin^2αsin^2β+cos^2αcos^2β-1/2cos2αcos2β 化简：sin²αsin²β+cos²αcos²β-1/2cos2αcos2β 化简 sin²αsin²β+cos²αcos²β-1/2（cos2αcos2β） sin²αsin²β+cos²αcos²β-【(1/2)*cos2αcos2β】 化简：sin²αsin²β+cos²αcos²β—1/2cos2αcos2β 化简sinα^2sinβ^2+cos^2cosβ^2－1/2cos2αcos2β 已知cos2α-cos2β=1/4,那么sin(α+β)sin(α-β)=? 化简:(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β