tanx2/tanx1〉x2/x1,0〈x1〈x2〈π/2

Question

tanx2/tanx1〉x2/x1 ,0〈x1〈 x2〈π/2

Answer 1

设函数f(x)=tanx/x,f'(x)=(xsec^2x-tanx)/x^2=g(x)/x^2
g'(x)=sec^2x+2xsec^2xtanx-sec^2x=2xsec^2xtanx>0
所以g(x)>g(0)=0,f'(x)>0,f(x)是增函数,所以
tanx2/x2>tanx1/x1
故tanx2/tanx1>x2/x1