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,
$\tan [{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)$
$+\cdots +{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]$ 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, $a_1,a_2,a_3,a_4\dots a_n$ is an arithmetic progression.
Then, $d=a_2-a_1=a_3-a_2=\cdots =a_n-a_{n-1}$
$\therefore \tan [{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)$
$+{\tan }^{-1}\left(\frac{d}{1+a_3{a}_4}\right)+\dots .+{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]$
$=\tan [{\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)$
$+{\tan }^{-1}\left(\frac{a_4-a_3}{1+a_3{a}_4}\right)+\dots +{\tan }^{-1}\left(\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n}\right)]$
$=\tan [{\tan }^{-1}{a}_2-{\tan }^{-1}{a}_1+{\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2+{\tan }^{-1}{a}_4$
$-{\tan }^{-1}{a}_3+\dots .+{\tan }^{-1}{a}_n-{\tan }^{-1}{a}_{n-1}]$
$=\left(\because {\tan }^{-1}\left[\frac{x-y}{1+xy}\right]={\tan }^{-1}x-{\tan }^{-1}y\right)$
$=\tan \left[{\tan }^{-1}{a}_n-{\tan }^{-1}{a}_1\right]$
$=\tan \left[{\tan }^{-1}\left(\frac{a_n-a_1}{1+a_n{a}_1}\right)\right]=\frac{a_n-a_1}{1+a_1{a}_n}$