Question

$\forall n\in N$, x $\in R$, the value of
${\tan }^{-1}\left[\frac{x}{1.2+x^2}\right]+{\tan }^{-1}\left[\frac{x}{2.3+x^2}\right]+\cdots +{\tan }^{-1}\left[\frac{x}{n(n+1)+x^2}\right]$, is

• A. ${\tan }^{-1}\left[\frac{x}{n}\right]-{\tan }^{-1}\left[\frac{x}{n+1}\right]$
• B. ${\tan }^{-1}\left[x\right]-{\tan }^{-1}\left[\frac{x}{n+1}\right]$
• C. ${\tan }^{-1}[n+1]-{\tan }^{-1}\left[x\right]$
• D. $\tan {}^{-1}\left[x\right]$

Answer 1

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

Let's assume $r^{th}$ term as

$t_r={\tan }^{-1}\left(\frac{x}{r.(r+1)+x^2}\right)={\tan }^{-1}\left(\frac{\frac{r+1}{x}-\frac{r}{x}}{\frac{r}{x}.\frac{r+1}{x}+1}\right)$

We know that,

${\tan }^{-1}\left(\frac{A-B}{1+AB}\right)={\tan }^{-1}A-{\tan }^{-1}B$

Thus,
$t_r={\tan }^{-1}\left(\frac{r+1}{x}\right)-{\tan }^{-1}\left(\frac{r}{x}\right)$

Now add by Difference Method varying $r$ from $1$ to $n$, we get
$S={\tan }^{-1}\left(\frac{n+1}{x}\right)-{\tan }^{-1}\left(\frac{1}{x}\right)$

Converting to ${\cot }^{-1}$ terms using ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$, we get

$S={\cot }^{-1}\left(\frac{1}{x}\right)-{\cot }^{-1}\left(\frac{n+1}{x}\right)$

Now, converting back to ${\tan }^{-1}$ terms by using ${\tan }^{-1}\left(\frac{1}{x}\right)={\cot }^{-1}x$, we get

$S={\tan }^{-1}x-{\tan }^{-1}\left(\frac{x}{n+1}\right)$

Hence, option B.