y=(sinx+tanx)/(cosx+1/tanx)
=sinx(1+1/cosx)/[cosx(1+1/sinx)]
=tanx*sinx(cosx+1)/[cosx(sinx+1)]
=tan^2x*(cosx+1)/(sinx+1)
由sinx,cosx∈[-1,1],tanx存在且不为0,可得出cosx+1>0,sinx+1>0
所以y>0