f(x)=tanx+tan(x+3/2π)……急
tanx+tan(π/2 -x)=?
急求函数y=tan²x-2tanx+3,(π/6≤x≤π)的最值
证明sec x+tanx=tan(π/4 +x/2)
tan(x/2)=2.求tanx
证明tanx+1/cosx=tan(x/2+π/4)
已知X属于[-π/3,π/4]求函数y=tan平方X +2tanx
f(x)=(2tanx/2)/(1-tan^2 x/2)
f(x)=tanx/√(1+tan^2x)最小正周期
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