Recall that,
#arctanx-arctany=arctan{(x-y)/(1+xy)}; x,y > 0........(star)#
Now, #f(x)=arctan(1/(x^2+x+1))=arctan{((x+1)-x)/(1+x(x+1))}#
#:.," by "(star), f(x)=arctan(x+1)-arctanx,.......(ast)#
#rArr A=f(1)+f(2)+f(3)+...+f(14)+f(15)#
#={arctan2-arctan1}+{arctan3-arctan2}+{arctan4-arctan3}+...
+{arctan15-arctan14}+{arctan16-arctan15},#
#:. A=arctan16-arctan1#
#=arctan{(16-1)/(1+16*1)},....[because, (star)]#
#:. A=arctan(15/17)#
#rArr tanA=15/17.#
Enjoy Maths.!