已知1-tana/2+tana=1 求证:tan2a=-4tan(a+π/4)

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已知1-tana/2+tana=1求证:tan2a=-4tan(a+π/4)已知1-tana/2+tana=1求证:tan2a=-4tan(a+π/4)已知1-tana/2+tana=1求证:tan2

已知1-tana/2+tana=1 求证:tan2a=-4tan(a+π/4)
已知1-tana/2+tana=1 求证:tan2a=-4tan(a+π/4)

已知1-tana/2+tana=1 求证:tan2a=-4tan(a+π/4)
1-tana/2+tana=1
1-tana=2+tana
tana=-1/2
所以tan2a=2tana/(1-tan²a)=-4/3
tan(A+π/4)=(tanA+tanπ/4)/(1-tanAtanπ/4)=(-1/2+1)/(1-(-1/2))=1/3
所以tan2a=-4tan(a+π/4)