Question

If $S_n={\cot }^{-1}\left(\frac{5}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{9}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{15}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$ upto $n$ terms, then

• A. $\tan \left(S_{10}\right)<1$
• B. $\tan \left(S_{10}\right)>1$
• C. $S_{\infty }=\frac{\pi }{3}$
• D. $S_{\infty }=\frac{\pi }{6}$

Answer 1

Answer: (A), (C)

$S_{\infty }=\frac{\pi }{3}$

$S_n={\cot }^{-1}\left(\frac{5}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{9}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{15}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{23}{\sqrt{3}}\right)+\cdots$
$\Rightarrow S_n={\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)+{\tan }^{-1}\left(\frac{\sqrt{3}}{9}\right)+{\tan }^{-1}\left(\frac{\sqrt{3}}{15}\right)+{\tan }^{-1}\left(\frac{\sqrt{3}}{23}\right)+\cdots$
$T_n={\tan }^{-1}\left\{\frac{\sqrt{3}}{n^2+n+3}\right\}={\tan }^{-1}\left\{\frac{\frac{n+1}{\sqrt{3}}-\frac{n}{\sqrt{3}}}{1+\frac{(n+1)}{\sqrt{3}}\cdot \frac{n}{\sqrt{3}}}\right\}$
$\Rightarrow T_n={\tan }^{-1}\left(\frac{n+1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{n}{\sqrt{3}}\right)$

$S_n=T_1+T_2+T_3+...+T_n$
$={\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(\frac{3}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)⋮⋮+{\tan }^{-1}\left(\frac{n+1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{n}{\sqrt{3}}\right)$

$\Rightarrow S_n={\tan }^{-1}\left(\frac{n+1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\therefore S_{10}={\tan }^{-1}\frac{11}{\sqrt{3}}-{\tan }^{-1}\frac{1}{\sqrt{3}}={\tan }^{-1}\left(\frac{10\sqrt{3}}{14}\right)$
$\Rightarrow \tan \left(S_{10}\right)=\frac{10\sqrt{3}}{14}>1$
and $S_{\infty }={\tan }^{-1}\left(\infty \right)-\frac{\pi }{6}=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$