Partial Fraction Decomposition?

We have two angles separated by $1^\circ$ in the denominator, so this might be helpful.

$\lim_{n\rightarrow1337}\sum_{i=1}^n\dfrac{1}{\cos i^{\circ}\times\cos(i-1)^{\circ}}\\ =\lim_{n\rightarrow1337}\sum_{i=1}^n\dfrac{\sin1^{\circ}}{\cos i^{\circ}\times\cos(i-1)^{\circ}\times\sin1^{\circ}}$

Let's take a look at a way to decompose this fraction by recalling a familiar identity.

$\tan x-\tan y\\ =\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}\\ =\dfrac{\sin x\times\cos y-\sin y\times\cos x}{\cos x\cos y}\\ =\dfrac{\sin(x-y)}{\cos x\cos y}$

In the problem, $x=i^{\circ}$ and $y=(i-1)^{\circ}\text{,}$ so we get this.

$\dfrac{\sin(x-y)^{\circ}}{\cos x^{\circ}\cos y^{\circ}}\\ =\dfrac{\sin(i-(i-1))^{\circ}}{\cos i^{\circ}\cos(i-1)^{\circ}}\\ =\dfrac{\sin1^{\circ}}{\cos i^{\circ}\cos(i-1)^{\circ}}$

So the sum is equal to this. $\lim_{n\rightarrow1337}\sum_{i=1}^n\dfrac{\sin1^{\circ}}{\cos i^{\circ}\times\cos(i-1)^{\circ}\times\sin1^{\circ}}\\ =\lim_{n\rightarrow1337}\sum_{i=1}^n\dfrac{\tan i^{\circ}-\tan(i-1)^{\circ}}{\sin1^{\circ}}$

Obviously, this telescopes, so the final answer is equal to this.

$\dfrac{\tan1337^{\circ}-\tan0^{\circ}}{\sin1^{\circ}}=\dfrac{\tan1337^{\circ}}{\sin1^{\circ}} =\dfrac{\tan77^{\circ}}{\sin1^{\circ}}=248.187\ldots$

The final answer is $2+4+8+1+8+7=\boxed{30}$