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Answer 1 · 2017-04-17T14:36:29

#15/17.#

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.!