Twelve? Ain't Nobody Got Time For That!

There is another way. Credits to my friend Srikanth for this. Let $re^{ix} = 1+2i$ [ Observe that $x=\tan^{-1}2$ ]. Raise both sides to power 12 and when you get the final complex number, say $a+ib$ , merely read off $b/a$ . It is easier to manipulate complex numbers than apply the tan formula.

We apply the compound angle formula repeatedly: $\tan (A+B) = \frac { \tan(A) + \tan(B)} {1 - \tan(A) \tan(B)}$ , so $\tan(2A) = \frac { 2\tan(A)} {1 - \tan^2 (A) }$

$\begin{aligned} \tan (2x) & = & \frac { 2 \cdot 3 }{1 - 2^2} = \frac {4}{3} \\ \tan (4x) & = & \frac { 2 \cdot \frac {4}{3} }{1 - \left ( \frac {4}{3} \right )^2} = \frac {24}{7} \\ \tan (8x) & = & \frac { 2 \cdot \frac {24}{7} }{1 - \left ( \frac {24}{7} \right )^2} =- \frac {336}{527} \\ \tan (12x) & = & \frac { \tan(8x) + \tan(4x) }{1 - \tan(8x) \tan(4x) } \\ & = & \frac { - \frac {336}{527} + \frac {24}{7} } {1 + \frac {336}{527} \cdot \frac {24}{7} } \\ & = & \frac {10296}{11753} \\ \end{aligned}$

$\Rightarrow (a+b) \pmod{1000} = \boxed{49}$

my question is that as we know tan(12x)=a/b and here we have tan(12x)=10296/11753, by comparing these two conditions we have a/b=10296/11753.so here a=10296 and b=11753 now we have to find a and b are coprime and then a+b.if we add aand b then we get 22049 and here we can tell last three digits of a+b is 049.that is correct.but i can't understand why (a+b) (mod 1000)? as we know coprime means gcd of a and will be 1 then why we divide (a+b) by 1000 and remainder is 49? my question is that why we take 1000?