y'=sec^2x,于是y'|(x=pi/4)=2切线方程为y-1=2(x-pi/4),即：2x-y+1-pi/2=0设法线方程为x+2y+C=0,由于它过M点1+pi/2+C=0,故C=-1-pi/2所以法线方程为x+2y-1-pi/2=0
