已知sin(a+b)=1 则tan(2a+b)+tanb
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已知sin(a+b)=1则tan(2a+b)+tanb已知sin(a+b)=1则tan(2a+b)+tanb已知sin(a+b)=1则tan(2a+b)+tanb4tan(a/2)=1-tan(a/2
已知sin(a+b)=1 则tan(2a+b)+tanb
已知sin(a+b)=1 则tan(2a+b)+tanb
已知sin(a+b)=1 则tan(2a+b)+tanb
4tan(a/2)=1-tan(a/2)得 2tan(a/2)/(1-tan(a/2))=1/2
tan(a)=1/2
4sinb=sin(2a+b)+sinb=2sin(a+b)cos(a)
2sinb=sin(2a+b)-sinb=2cos(a+b)sin(a)
相除得2tan(a)=tan(a+b)=1 a+b=π/4
已知sin(a+b)=1 则tan(2a+b)+tanb
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