Question

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

• A. $\frac{5}{11}$
• B. $\frac{5}{6}$
• C. $-\frac{6}{5}$
• D. $\frac{10}{11}$

Answer 1

The correct option is B

$\frac{5}{6}$

Explanation for the correct option.

Step 1: Using this foumula, ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$.

Given, $\begin{array}{rcl}\mathrm{S}& =& {\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{13}\right)+\dots \end{array}$

Now using, ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$.

$\begin{array}{rcl}\therefore \mathrm{S}& =& {\tan }^{-1}\left(\frac{2-1}{1+1.2}\right)+{\tan }^{-1}\left(\frac{3-2}{1+2\times 3}\right)+{\tan }^{-1}\left(\frac{4-3}{1+3\times 4}\right)+\dots .+{\tan }^{-1}\left(\frac{11-10}{1+10\times 11}\right)\\ & =& \left({\tan }^{-1}2-{\tan }^{-1}1\right)+\left({\tan }^{-1}3-{\tan }^{-1}2\right)+\left({\tan }^{-1}4-{\tan }^{-1}3\right)+\dots .+\left({\tan }^{-1}\left(11\right)-{\tan }^{-1}\left(10\right)\right)\\ & =& {\tan }^{-1}11-{\tan }^{-1}1\\ & =& {\tan }^{-1}\left(\frac{11-1}{1+11}\right)\end{array}$]

Step 2: Finding the $\tan \left(S\right)$

$\begin{array}{rcl}\therefore \tan \left(S\right)& =& \frac{11-1}{1+11\times 1}\\ & =& \frac{5}{6}\end{array}$

Hence, option B is correct.