Question

If $a_1,a_2,a_3,...,a_n$ is an arithmetic progression with common difference d, then evaluate the following expression.

$tan[ta{n}^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+ta{n}^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+ta{n}^{-1}\left(\frac{d}{1+a_3{a}_4}\right)+\cdots +ta{n}^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]$

Answer 1

We have, $a_1=a,a_2=a+d,a_3=a+2d$

and $d=a_2-a_1=a_3-a_2=a_4-a_3=\cdots =a_n-a_{n-1}$

Given that, $tan[ta{n}^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+ta{n}^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+ta{n}^{-1}\left(\frac{d}{1+a_3{a}_4}\right)+\cdots +ta{n}^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]$

$=tan[ta{n}^{-1}\frac{a_2-a_1}{1+a_2.a_1}+ta{n}^{-1}\frac{a_3-a_2}{1+a_3.a_2}+\cdots +ta{n}^{-1}\frac{a_n-a_{n-1}}{1+a_n.a-n-1}]=tan\left[\right(ta{n}^{-1}{a}_2-ta{n}^{-1}{a}_1)+(ta{n}^{-1}{a}_3-ta{n}^{-1}{a}_2)+\cdots +(ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_{n-1}\left)\right]=tan[ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_1]=tan\left[ta{n}^{-1}\frac{a_n-a_1}{1+a_n.a_1}\right][\because ta{n}^{-1}x-ta{n}^{-1}y-=ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)]=\frac{a_n-a_1}{1+a_n.a_1}[\because tan(ta{n}^{-1}x)=x]$