Incredible Tangential Sum

Here is the general formula for this type of sum:

$\tan \Bigg(\displaystyle\sum_{n=1}^x \tan^{-1} n\Bigg) = \frac{\large Im\Bigg[\displaystyle\prod_{n=1}^x (1+ni)\Bigg]}{\large Re \Bigg[\displaystyle\prod_{n=1}^x (1+ni)\Bigg]}$

Basically what this formula is saying that you have to find the product $(1+i)\times(1+2i)\times \dots \times (1+xi)$ , and your answer is the imaginary part of this number divided by the real part.

For this specific question, our product terminates to $5864300+3103100 i$ . Taking the imaginary part and dividing by the real part simplifies to $\frac{2387}{4511}$ , therefore our answer is $2387+4511=\boxed{6898}$ .

For a proof of this formula, please refer to this note .

First, we will prof that $\tan^{-1} A+\tan^{-1} B=\tan^{-1}\left(\frac{A+B}{1-AB}\right)$

Using the identity $\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}\implies X+Y=\tan^{-1}\left(\frac{\tan X+\tan Y}{1-\tan X\tan Y}\right)$ Let $X=\tan^{-1} A$ and $Y=\tan^{-1} B$ and substitute to the last equation and the evidence has been obtained.

Now, we will use the formula to evaluate each of the two arc-tangent addition.

(1) $~~\tan^{-1}(1)+\tan^{-1}(2)=\tan^{-1}(-3)$

(2) $~~\tan^{-1}(3)+\tan^{-1}(4)=\tan^{-1}(-\frac{7}{11})$

(3) $~~\tan^{-1}(5)+\tan^{-1}(6)=\tan^{-1}(-\frac{11}{29})$

(4) $~~\tan^{-1}(7)+\tan^{-1}(8)=\tan^{-1}(-\frac{3}{11})$

(5) $~~\tan^{-1}(9)+\tan^{-1}(10)=\tan^{-1}(-\frac{19}{89})$

furthermore, add (1) and (2), (3) and (4)

(6) $~~\tan^{-1}(-3)+\tan^{-1}(-\frac{7}{11})=\tan^{-1}\left(\frac{-3-\frac{7}{11}}{1-\frac{21}{11}}\right)=\tan^{-1}(4)$

(7) $~~\tan^{-1}(-\frac{11}{29})+\tan^{-1}(-\frac{3}{11})=\tan^{-1}(-\frac{8}{11})$

Now, add (6) and (7)

(8) $~~\tan^{-1}(4)+\tan^{-1}(-\frac{8}{11})=\tan^{-1}(\frac{36}{43})$

The last, add (5) and (8)

(9) $~~\tan^{-1}(-\frac{19}{89})+\tan^{-1}(\frac{36}{43})=\tan^{-1}(\frac{2387}{4511})$

Finally, the answer of our problem is

$\tan \left[\tan^{-1}\left(\frac{2387}{4511}\right)\right]=\frac{2387}{4511}=\frac{a}{b}$

Thus, $a+b=\boxed{\boxed{\boxed{\LARGE{6898}}}}$

Note: this is not a practical solution. I'll wait for another more practical solution