求函数值的题

设arctan1=t,arctan2=s,arctan3=k,则tant=1,tan s=2,tan k=3,tan(s+k)=(2+3)/(1-2*3)=-1,于是,tan(t+s+k)=(tant+tan(s+k))/(1-tant*tan(s+k))tan(t+s+k)=0,t+s+k=n*PI(n属于整数）又因为,arctan(x)为增函数,所以arct...