The year in question .....

Note first that, in general,

$\cot(x) - \cot(2x) =\dfrac{\cos(x)}{\sin(x)} - \dfrac{\cos(2x)}{\sin(2x)} =$

$\dfrac{\cos(x)}{\sin(x)} - \dfrac{\cos(2x)}{2\sin(x)\cos(x)} = \dfrac{2\cos^{2}(x) - \cos(2x)}{2\sin(x)\cos(x)} =$

$\dfrac{2\cos^{2}(x) - (2\cos^{2}(x) - 1)}{\sin(2x)} = \dfrac{1}{\sin(2x)}$ .

So $\csc((2^{m})^{\circ}) = \cot((2^{m-1})^{\circ}) - \cot((2^{m})^{\circ})$ . Thus when we expand $S$ we end up with a telescoping series, leaving us in the end with just

$\cot(4^{\circ}) - \cot((2^{2019})^{\circ})$ .

We now need to determine $2^{2019} \mod{360}$ . To this end we note that

$2^{15} \equiv 2^{3} \mod{360}$ ,

so for powers of $2$ of the form $p = 3 + 12*q$ for any integer $q$ we have that $2^{p} \equiv 8 \mod{360}$ . Since $2019 = 3 + 12*168$ , we now have that

$S = \cot(4^{\circ}) - \cot(8^{\circ}) = \dfrac{\cos(4^{\circ})}{\sin(4^{\circ})} - \dfrac{\cos(8^{\circ})}{\sin(8^{\circ})} =$

$\dfrac{\sin(8^{\circ})\cos(4^{\circ}) - \cos(8^{\circ})\sin(4^{\circ})}{\sin(4^{\circ})\sin(8^{\circ})} = \dfrac{1}{\sin(8^{\circ})} = \sec(82^{\circ}).$ Thus $k = \boxed{82}$ .

initialize with csc(x)=cot(x/2)-cot(x) then we are getting a telescoping sum which reduces to cot (4)-cot (2^2019) . to calculate 2^2019 mod(360) One can use the Chinese remainder theorem