Question #a9b54

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Answer 1 · 2017-04-17T14:36:29

$\frac{15}{17} .$ $15/17.$

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.!