If $$S$$ is the sum of the first 10 terms of the series, $$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \ldots$$, then $$\tan(S)$$ is equal to:

We have to find $$\tan(S)$$ where $$S$$ is the sum of the first ten angles of the series

$$\tan^{-1}\!\left(\frac{1}{3}\right)+\tan^{-1}\!\left(\frac{1}{7}\right)+\tan^{-1}\!\left(\frac{1}{13}\right)+\tan^{-1}\!\left(\frac{1}{21}\right)+\tan^{-1}\!\left(\frac{1}{31}\right)+\tan^{-1}\!\left(\frac{1}{43}\right)+\tan^{-1}\!\left(\frac{1}{57}\right)+\tan^{-1}\!\left(\frac{1}{73}\right)+\tan^{-1}\!\left(\frac{1}{91}\right)+\tan^{-1}\!\left(\frac{1}{111}\right).$$

The only tool we really need is the tangent-addition formula. We first state it clearly:

For any two angles $$A$$ and $$B$$,

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

Because every term in the given series is already an inverse-tangent, we can treat each of them as an angle whose tangent is the given fraction, and add them one by one, updating the tangent each time with the formula above.

Let us denote

$$T_1=\tan^{-1}\!\left(\frac{1}{3}\right),\; T_2=\tan^{-1}\!\left(\frac{1}{7}\right),\; T_3=\tan^{-1}\!\left(\frac{1}{13}\right),\; \ldots,\; T_{10}=\tan^{-1}\!\left(\frac{1}{111}\right).$$

Also let

$$S_k=T_1+T_2+\ldots+T_k\quad\text{and}\quad x_k=\tan(S_k).$$

We begin with the very first term.

For $$k=1$$ we have $$S_1=T_1,$$ so

$$x_1=\tan(S_1)=\tan\!\left(\tan^{-1}\!\left(\frac{1}{3}\right)\right)=\frac{1}{3}.$$ So $$x_1=\dfrac{1}{3}.$$

Now we incorporate the second term. Using the tangent-addition formula with $$A=S_1,\;\tan A=x_1=\frac13$$ and $$B=T_2,\;\tan B=\frac17$$, we get

$$x_2=\tan(S_2)= \frac{\dfrac13+\dfrac17}{1-\dfrac13\cdot\dfrac17} =\frac{\dfrac{10}{21}}{1-\dfrac{1}{21}} =\frac{\dfrac{10}{21}}{\dfrac{20}{21}} =\frac{10}{20} =\frac12.$$

Next we add the third term. Put $$A=S_2,\;\tan A=x_2=\frac12$$ and $$B=T_3,\;\tan B=\frac{1}{13}.$$ Again,

$$x_3=\frac{\dfrac12+\dfrac1{13}}{1-\dfrac12\cdot\dfrac1{13}} =\frac{\dfrac{15}{26}}{1-\dfrac1{26}} =\frac{\dfrac{15}{26}}{\dfrac{25}{26}} =\frac{15}{25} =\frac35.$$

We proceed similarly for the fourth term, taking $$\tan A=x_3=\dfrac35$$ and $$\tan B=\dfrac1{21}$$:

$$x_4=\frac{\dfrac35+\dfrac1{21}}{1-\dfrac35\cdot\dfrac1{21}} =\frac{\dfrac{68}{105}}{1-\dfrac3{105}} =\frac{\dfrac{68}{105}}{\dfrac{102}{105}} =\frac{68}{102} =\frac23.$$

Adding the fifth term, with $$\tan A=x_4=\dfrac23$$ and $$\tan B=\dfrac1{31}$$, gives

$$x_5=\frac{\dfrac23+\dfrac1{31}}{1-\dfrac23\cdot\dfrac1{31}} =\frac{\dfrac{65}{93}}{1-\dfrac2{93}} =\frac{\dfrac{65}{93}}{\dfrac{91}{93}} =\frac{65}{91} =\frac57.$$

For the sixth term we have $$\tan A=x_5=\dfrac57$$ and $$\tan B=\dfrac1{43}$$, so

$$x_6=\frac{\dfrac57+\dfrac1{43}}{1-\dfrac57\cdot\dfrac1{43}} =\frac{\dfrac{222}{301}}{1-\dfrac5{301}} =\frac{\dfrac{222}{301}}{\dfrac{296}{301}} =\frac{222}{296} =\frac{37\cdot6}{37\cdot8} =\frac34.$$

With the seventh term, $$\tan A=x_6=\dfrac34$$ and $$\tan B=\dfrac1{57},$$ so

$$x_7=\frac{\dfrac34+\dfrac1{57}}{1-\dfrac34\cdot\dfrac1{57}} =\frac{\dfrac{175}{228}}{1-\dfrac3{228}} =\frac{\dfrac{175}{228}}{\dfrac{225}{228}} =\frac{175}{225} =\frac79.$$

Adding the eighth term, $$\tan A=x_7=\dfrac79$$ and $$\tan B=\dfrac1{73},$$ yields

$$x_8=\frac{\dfrac79+\dfrac1{73}}{1-\dfrac79\cdot\dfrac1{73}} =\frac{\dfrac{520}{657}}{1-\dfrac7{657}} =\frac{\dfrac{520}{657}}{\dfrac{650}{657}} =\frac{520}{650} =\frac{52}{65} =\frac45.$$

For the ninth term, with $$\tan A=x_8=\dfrac45$$ and $$\tan B=\dfrac1{91},$$ we have

$$x_9=\frac{\dfrac45+\dfrac1{91}}{1-\dfrac45\cdot\dfrac1{91}} =\frac{\dfrac{369}{455}}{1-\dfrac4{455}} =\frac{\dfrac{369}{455}}{\dfrac{451}{455}} =\frac{369}{451} =\frac{9\cdot41}{11\cdot41} =\frac{9}{11}.$$

Finally we add the tenth term: $$\tan A=x_9=\dfrac{9}{11}$$ and $$\tan B=\dfrac1{111}.$$ Thus

$$x_{10}=\frac{\dfrac{9}{11}+\dfrac1{111}}{1-\dfrac{9}{11}\cdot\dfrac1{111}} =\frac{\dfrac{1010}{1221}}{1-\dfrac9{1221}} =\frac{\dfrac{1010}{1221}}{\dfrac{1212}{1221}} =\frac{1010}{1212} =\frac{505}{606} =\frac{5}{6}.$$

After ten terms we have reached $$S=S_{10}$$, and the tangent of this total angle is therefore

$$\boxed{\dfrac{5}{6}}.$$

Hence, the correct answer is Option A.