求值:tan(2π+π/4 )+tan(4π+π/4)+tan(6π+π/4).+tan(2010π+π/4)=
tan(a+π4
化简:{[tan(π/4+a)-tan(π/4-a)]/tan2a} *tan(π/4+a)
tan(a+b)=3/4,tan(a-π/4)=1/2,那么tan(b+π/4)等于多少
反函数求值tan[arcsin(-1/4)+π/3]
关于三角函数恒等变化1-1/2tanα-1/2tan(π/4-α)=2-1/2[(1+tanα+1/(1+tanα)]
求值tan(π/6-θ)+tan(π/6+θ)+√3tan(π/6-θ)tan(π/6+θ)
已知tan a=3,求tan(a+π/4),tan(a-π/4)的值?
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
求证:tan(a+π/4)+tan(a-π/4)=2tan2α