不是,解答如下：设：直角三角形三边长对应的分别为a,b,c(x2+y2=r2).再设a=x,b=y,c=r则,cosx=x/r,secx=r/x.所以,cos2x=cosx2-sinx2=x2/r2-y2/r2=(x2-y2)/r2.而sec2x=r2/x2即cos2x*sec2x=[(x2-y2)/r2]*(r2/x2)=(x2-y2)...
