y=tan(π/4-x)为什么又等于-tan(x-π/4)
高一数学y=tan(x+π/4)+1/tan(x+π/4)
函数y=tan(π/2-x)(-π/4
y=3tan(2x-π/4)
求y=tan(x/2 -π/4)的周期
设tan[x+y]=5分之2,tan[y-4分之π,求]tan[x+4分之π的值]
函数y=tan(x-π6
若tan(x)=4tan(y),求x,y关系 (已知0
1)tan(x/2+π/4)+tan(x/2-π/4)=2tanx
tan(X/2+π/4)+tan(x/2-π/4)=2tanx?
tan( x/2+π/4)+tan(x/2-π/4 )=2tanx
tan(x/2+ π4)+tan(x/2- π/4)=2tanx
tan(π/4 - x)为什么等于(1-tanx)/(1+tanx)?