tan(x-y)=(tanx-tany)/(1+tanxtany)=2tany/(1+3tan²y)
已知0＜x,y＜π/2,tanx＝3tany.求证x-y≤π/6
1个回答
相关问题
-
如果实数x、y满足|tanx|+|tany|＞|tanx+tany|，且y∈(π，3π2)，则|tanx-tany|等于
-
如果实数x、y满足|tanx|+|tany|＞|tanx+tany|，且y∈(π，3π2)，则|tanx-tany|等于
-
x+y=3π/4,则(1-tanx)(1-tany)
-
若x=π/16,y=3π/16 ,则（1+tanx)(1+tany)=?
-
求证tanx+tany/tanx-tany=sin(x+y）/sin(x-y)
-
已知tanx=2,tany=3,x,y均为锐角,求证：x+y=135°
-
已知X,Y,Z∈R,X+Y+Z=π,X²+Y²+Z²=π²/2,求证π2／3≥X≥0,π2／3≥Y≥0,π2／3≥Z≥0
-
若0≤x-y＜π/2,且tanx=3tany,则x-y的最大值为?
-
已知x.y∈(0,π）且tan(x-y）=1/2,tany=-1/7,求2x-y
-
已知tanx,tany是方程x^2+6x+7=0的两个根,求证sin(x+y)=cos(x+y).