若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz
若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz成立.用此结论来证明恒等式
已知x,y,z都是锐角,sin^2x+sin^2y+sin^2z=1,求tanx*tany*tanz的最值
求证tanx+tany/tanx-tany=sin(x+y)/sin(x-y)(详细步骤)
证明 (tanX+tanY)/(tanX-tanY)=(sin(X+Y))/(sin(X-Y))
已知sin(x+y)/cos(x-y)=m/n,则tanx/tany=?
已知tanx=2,tany=3,x,y∈(0,π/2),求x+y.
一道数奥题 解方程组 (tanx)^2+2(cot2y)^2=1(tany)^2+2(cot2z)^2=1(tanz)^
已知tanx,tany是方程x^2+6x+7=0的两个根,求证sin(x+y)=cos(x+y).
若sin(x+y)=1/2,sin(x-y)=1/3,则tanx/tany等于?
若x、y是锐角且tanx=2/3tany=3/4则sin(x+y)等于
cos(x+y)=1/5,cos(x-y)=3/5则tanx.tany=?
已知sinx+siny+sinz=cosx+cosy+cosz=0,求tan(x+y+z)+tanxtanytanz的值