1）sinx-cosx=√2[sin(x-π/4)],x-π/4∈(-π/4,3π/4),
显然当x∈（0,π/4）时,√2sin(x-π/4）cosx
2)tanx-cosx=[sinx-(cosx)^2]/cosx=[(sinx)^2+sinx-1]/cosx.
x∈（0,π/2）∪（π/2,π),0