1. Identify the general term

The denominators are: $3,7,13,21,\dots$ These follow the pattern $n(n+1)+1$ for $n=1,2,3,4,\dots$, since: $1\cdot 2+1=3,\quad 2\cdot 3+1=7,\quad 3\cdot 4+1=13,\quad 4\cdot 5+1=21$

So the $n$th term is $\tan^{-1}\left(\frac{1}{n(n+1)+1}\right).$

Hence $S=\sum_{n=1}^{10} \tan^{-1}\left(\frac{1}{n(n+1)+1}\right).$

2. Rewrite each term using an arctangent identity

We use: $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$ when the angles lie in the principal range.

Take $a=n+1,\qquad b=n.$ Then $\tan^{-1}(n+1)-\tan^{-1}(n)=\tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)=\tan^{-1}\left(\frac{1}{n(n+1)+1}\right).$

So each term becomes $\tan^{-1}\left(\frac{1}{n(n+1)+1}\right)=\tan^{-1}(n+1)-\tan^{-1}(n).$

3. Form the telescoping sum

Therefore, $S=\sum_{n=1}^{10}\Big(\tan^{-1}(n+1)-\tan^{-1}(n)\Big).$

Expanding: $S=(\tan^{-1}2-\tan^{-1}1)+(\tan^{-1}3-\tan^{-1}2)+\cdots +(\tan^{-1}11-\tan^{-1}10).$

All intermediate terms cancel, leaving $S=\tan^{-1}(11)-\tan^{-1}(1).$

Since $\tan^{-1}(1)=\frac{\pi}{4},$ we get $S=\tan^{-1}(11)-\frac{\pi}{4}.$

4. Compute $\tan S$

Using $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B},$ with $A=\tan^{-1}(11),\quad B=\frac{\pi}{4},$ we have $\tan A=11,\qquad \tan B=1.$

Thus $\tan S=\frac{11-1}{1+11\cdot 1} = \frac{10}{12}=\frac{5}{6}.$

5. Check options

The value is $\tan(S)=\frac{5}{6}.$ So the correct option is:

D: $\frac{5}{6}$