tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=
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tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=tan(a+π/4)=2,则cos2a+3sin^2a+tan2
tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=
tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=
tan(a+π/4)=2,则cos2a+3sin^2a+tan2a=
先把已知条件展开,[tana+tan(π/4)]/[1-tanatan(π/4)]=2 →( tana+1)/(1-tana)=2,可以解出tana1/3.
然后把要求的式子化简：原式=1-2sin^2a+3sin^2a+tan2a=sin^2a+tan2a
已知tan2a=2tana/(1-tan^2a)=3/4.
由tana=1/3,得 sina/cosa=1/3→3sina=cosa→又只sin^2a+cos^2a=1,代入3sina=cosa,得：sin^2a=1/10.
所以可得原式=3/4+1/10=17/20
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