已知sin(x+y)=1,求证:tan(2x+y)+tany=0
已知sin(x+y)=1,求证:tan(2x+y)+tany=0
已知sin(x+y)=1,求证:tan(2x+y)+tany=0
已知sin(x+y)=1,求证:tan(2x+3y)=tany
sin(x+y)=1\2,sin(x—y)=1\3,求[tan(x+y)-tanx-tany]\[tany的平方tan(
sin(x+y)=1,则tan(2x+y)+tany=
已知tanx,tany是方程x^2+6x+7=0的两个根,求证sin(x+y)=cos(x+y).
已知2cos(2x+y)=cosy,求tan(x+y)tany的值
求证tanx+tany/tanx-tany=sin(x+y)/sin(x-y)(详细步骤)
已知x.y∈(0,π)且tan(x-y)=1/2,tany=-1/7,求2x-y
证明 (tanX+tanY)/(tanX-tanY)=(sin(X+Y))/(sin(X-Y))
已知tanx=1/4,tany=-3,求tan(x+y)的值
已知tanx=-3/4,且tan(x+y)=1,求tany的值