Question

​​​​​The sum of $n$ terms of the series
${\text{cosec}}^{-1}\sqrt{10}+{\text{cosec}}^{-1}\sqrt{50}+{\text{cosec}}^{-1}\sqrt{170}+\cdots +{\text{cosec}}^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$ is

• A. ${\cot }^{-1}\left[\frac{n}{n+2}\right]$
• B. ${\tan }^{-1}\left[\frac{n}{n+2}\right]$
• C. ${\sin }^{-1}\left[\frac{n+2}{n}\right]$
• D. ${\sin }^{-1}\left[\frac{n}{3n^2+3n+4}\right]$

Answer 1

The correct option is B ${\tan }^{-1}\left[\frac{n}{n+2}\right]$

Let $\theta ={\text{cosec}}^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$
$\Rightarrow \text{cosec}\theta =\sqrt{(n^2+1)(n^2+2n+2)}$
$\Rightarrow {\text{cosec}}^2\theta =(n^2+1)(n^2+2n+2)$
$=n^4+2n^3+3n^2+2n+2$
$=(n^2+n+1{)}^2+1$
$\Rightarrow {\text{cosec}}^2\theta -1=(n^2+n+1{)}^2$
$\Rightarrow {\cot }^2\theta =(n^2+n+1{)}^2$
$\Rightarrow \tan \theta =\frac{1}{n^2+n+1}=\frac{(n+1)-n}{1+(n+1)n}$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{(n+1)-n}{1+(n+1)n}\right]$
$\Rightarrow \theta ={\tan }^{-1}(n+1)-{\tan }^{-1}n$

Thus, sum of $n$ terms of the given series
$=({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)+({\tan }^{-1}4-{\tan }^{-1}3)+\cdots +\left({\tan }^{-1}\right(n+1)-{\tan }^{-1}n)$
$={\tan }^{-1}(n+1)-{\tan }^{-1}1$
$={\tan }^{-1}\left[\frac{n+1-1}{1+(n+1)\cdot 1}\right]={\tan }^{-1}\left[\frac{n}{n+2}\right]$

Alternate Solution:
Put the value of $n$ as $1$ and verify the options.