1.证明tanx+tany=tan(x+y)-tanxtanytan(x+y)
证明 (tanX+tanY)/(tanX-tanY)=(sin(X+Y))/(sin(X-Y))
sin(x+y)=1\2,sin(x—y)=1\3,求[tan(x+y)-tanx-tany]\[tany的平方tan(
tanx+tany=25 cotx+coty=30,那么tan(x+y)=?
tanx+tany=25,cotx+coty=30.tan(x+y)=?
tanX=2分之1,tan(x+y)=5分之2,那么tanY=?
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
求证tanx+tany/tanx-tany=sin(x+y)/sin(x-y)(详细步骤)
已知tanx+tany=5m,tan(x+y)=6m(m≠0),tan(x-y)=1/7,求m的值及tanx,tany的
若x+y+z=nπ,求证:tanx+tany+tanz=tanxtanytanz成立.用此结论来证明恒等式
cos(x+y)=1/5,cos(x-y)=3/5则tanx.tany=?
已知tanx=2,tany=3,x,y∈(0,π/2),求x+y.