The value of $$\cot\left(\sum_{n=1}^{19} \cot^{-1}\left(1 + \sum_{p=1}^{n} 2p\right)\right)$$ is:

We have to evaluate the quantity

$$\cot\!\left(\;\sum_{n=1}^{19} \cot^{-1}\!\left(1+\sum_{p=1}^{n}2p\right)\right).$$

First let us simplify the inner summation. The sum of the first $$n$$ even natural numbers can be written as $$2$$ times the sum of the first $$n$$ natural numbers. Using the well-known formula

$$1+2+\dots +n=\frac{n(n+1)}{2},$$

we get

$$\sum_{p=1}^{n}2p \;=\;2\left(\frac{n(n+1)}{2}\right)=n(n+1).$$

So the expression inside each $$\cot^{-1}$$ becomes

$$1+\sum_{p=1}^{n}2p = 1+n(n+1)=n^2+n+1.$$

Therefore the entire angle whose cotangent is being taken is

$$S=\sum_{n=1}^{19}\cot^{-1}\!\bigl(n^2+n+1\bigr).$$

To telescope this sum, it is easier to convert $$\cot^{-1}$$ to $$\tan^{-1}$$. Recall that for any positive $$x$$ we have

$$\cot^{-1}x = \tan^{-1}\!\left(\frac1x\right).$$

Applying this to each term,

$$\cot^{-1}\!\bigl(n^2+n+1\bigr)=\tan^{-1}\!\left(\frac1{n^2+n+1}\right).$$

Now we invoke the subtraction formula for $$\tan^{-1}$$. For real numbers $$\alpha$$ and $$\beta$$,

$$\tan^{-1}\alpha-\tan^{-1}\beta=\tan^{-1}\!\left(\frac{\alpha-\beta}{1+\alpha\beta}\right).$$

If we choose $$\alpha=n+1$$ and $$\beta=n$$, then

$$\tan^{-1}(n+1)-\tan^{-1}n=\tan^{-1}\!\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}\!\left(\frac1{n^2+n+1}\right).$$

This is exactly the form we obtained above. Hence, for every $$n\ge1$$,

$$\cot^{-1}\!\bigl(n^2+n+1\bigr)=\tan^{-1}(n+1)-\tan^{-1}n.$$

Substituting this identity into the sum for $$S$$ gives

$$S=\sum_{n=1}^{19}\Bigl[\tan^{-1}(n+1)-\tan^{-1}n\Bigr].$$

Observe that this is a telescoping series: consecutive terms cancel pairwise.

Explicitly, writing out the first few terms,

$$\bigl[\tan^{-1}2-\tan^{-1}1\bigr]+\bigl[\tan^{-1}3-\tan^{-1}2\bigr] +\dots+\bigl[\tan^{-1}20-\tan^{-1}19\bigr].$$

All the intermediate $$\tan^{-1}2,\tan^{-1}3,\dots,\tan^{-1}19$$ terms cancel, leaving only

$$S=\tan^{-1}20-\tan^{-1}1.$$

Next we need $$\cot S$$. First find $$\tan S$$ using the subtraction formula for tangent:

For angles $$A$$ and $$B$$,

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B}.$$

Taking $$A=\tan^{-1}20$$ (so $$\tan A=20$$) and $$B=\tan^{-1}1$$ (so $$\tan B=1$$), we have

$$\tan S=\tan\!\bigl(\tan^{-1}20-\tan^{-1}1\bigr) =\frac{20-1}{1+20\cdot1} =\frac{19}{21}.$$

Finally, since $$\cot S$$ is the reciprocal of $$\tan S$$,

$$\cot S=\frac{1}{\tan S}=\frac{21}{19}.$$

Hence, the correct answer is Option A.