Question

If $a_1, a_2, a_3, \cdots, a_n$ is an arithmetic progression with common difference $d$, then the value of the expression,
$\tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)\right. +\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)$
$ \left.+\cdots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_n}\right)\right]$ is

• A. $a_n-a_n$
• B. $\frac{a_n-a_1}{a_1 a_n}$
• C. $\frac{a_n-a_1}{1+a_n a_n}$
• D. Does not exist

Answer 1

Answer: (C)

Given, $a_1, a_2, a_3, a_4 \ldots a_n$ is an arithmetic progression.
Then, $d=a_2-a_1=a_3-a_2=\cdots=a_n-a_{n-1}$
$\therefore \tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. $
$\left.+\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right)+\ldots .+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_n}\right)\right] $
$=\tan \left[\tan ^{-1}\left(\frac{a_2-a_1}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{a_3-a_2}{1+a_2 a_3}\right)\right.$
$\left.+\tan ^{-1}\left(\frac{a_4-a_3}{1+a_3 a_4}\right)+\ldots+\tan ^{-1}\left(\frac{a_n-a_{n-1}}{1+a_{n-1} a_n}\right)\right] $
$=\tan \left[\tan ^{-1} a_2-\tan ^{-1} a_1+\tan ^{-1} a_3-\tan ^{-1} a_2+\tan ^{-1} a_4\right. $
$\left. -\tan ^{-1} a_3+\ldots .+\tan ^{-1} a_n-\tan ^{-1} a_{n-1}\right]$
$=\left(\because \tan ^{-1}\left[\frac{x-y}{1+x y}\right]=\tan ^{-1} x-\tan ^{-1} y\right) $
$=\tan \left[\tan ^{-1} a_n-\tan ^{-1} a_1\right]$
$=\tan \left[\tan ^{-1}\left(\frac{a_n-a_1}{1+a_n a_1}\right)\right]=\frac{a_n-a_1}{1+a_1 a_n}$