✔ 最佳答案
設f(x)=tanx+sinx-2x, 0 < x < <π/2
f'(x)=(secx)^2+cosx -2
f"(x)=2(secx)^2*tanx-sinx= sinx*[2(secx)^3 -1]>0
f"(x)>0 , 則f'(x)遞增, so, f'(x)>f'(0)=0, 即f'(x)>0, for 0 < x < <π/2
f'(x)>0, 故 f(x)遞增, so, f(x)>f(0)= 0 for 0 < x < <π/2
f(x)>0, 即 tanx+sinx>2x for 0 < x < π/2