Question

lf $a_1,a_2,a_3\dots \dots ..a_n$ are in A.P. with common difference d, then $tan\left[ta{n}^{-1}\right(\frac{d}{1+a_1{a}_2})+ta{n}^{-1}(\frac{d}{1+a_2{a}_3})+\dots \dots \dots \dots \dots \dots +ta{n}^{-1}(\frac{d}{1+a_{n-1}{a}_n}\left)\right]=$ ?

• A. $\frac{(n-1)d}{a_1+a_n}$
• B. $\frac{(n-1)d}{1+a_1{a}_n}$
• C. $\frac{nd}{1+a_1{a}_n}$
• D. $\frac{a_n-a_1}{a_n+a_1}$

Answer 1

Answer: (B)

$\frac{(n-1)d}{1+a_1{a}_n}$

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