My 16th Birthday Problem

$sec^2(x) = tan^2(x)+1$ , $sec (x) = \sqrt{tan^2(x)+1}$

Let $a = tan (x)$ , then $arctan(a)=x$ , $sec (x) = sec (arctan(a))$ , $sec (arctan(a)) = f(a) = \sqrt{tan^2(x)+1} = \sqrt{a^2+1}$

So, $f(a) = \sqrt{a^2+1}$ , $f(f(a)) = \sqrt{a^2+2}$ . By repeatedly applying this, we get that $f(f(f(...a))) = \sqrt{a^2+n}$ where the function is iterated $n$ times.

So, our answer is $\sqrt{1+2015} = \sqrt{2016} = 12 \sqrt{14}$

So, our answer is $12+14 = \boxed{26}$