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:
Correct Option: 1
$S=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}+\ldots .$ upto 10 terms
$=\tan ^{-1}\left(\frac{2-1}{1+2 \cdot 1}\right)+\tan ^{-1}\left(\frac{3-2}{1+3 \cdot 2}\right)$
$+\tan ^{-1}\left(\frac{4-3}{1+3 \cdot 4}\right)+\ldots \ldots+\tan ^{-1}\left(\frac{11-10}{1+11 \cdot 10}\right)$
$=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+$
$\left(\tan ^{-1} 4-\tan ^{-1} 3\right)+\ldots . .+\left(\tan ^{-1} 11-\tan ^{-1} 10\right)$
$\operatorname{man}^{-1} 11-\tan ^{-1} 1=\tan ^{-1}\left(\frac{11-1}{1+11 \cdot 1}\right)=\tan ^{-1}\left(\frac{5}{6}\right)$
$\therefore \tan (S)=\frac{5}{6}$