Question

Let $a_1=1,a_2,a_3,a_4,\dots .$. be consecutive natural numbers. Then ${\tan }^{-1}\left(\frac{1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{1}{1+a_2{a}_3}\right)+\dots .+{\tan }^{-1}\left(\frac{1}{1+a_{2021}{a}_{2022}}\right)$ is equal to

• A. ${\cot }^{-1}(2022)-\frac{\pi }{4}$
• B. $\frac{\pi }{4}-{\cot }^{-1}(2022)$
• C. ${\tan }^{-1}(2022)-\frac{\pi }{4}$
• D. $\frac{\pi }{4}-{\tan }^{-1}(2022)$

Answer 1

Answer: (C)

$a_2-a_1=a_3-a_2=\dots ..=a_{2022}-a_{2021}=1$.
$\therefore {\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)+\dots ..+{\tan }^{-1}\left(\frac{a_{2022}-a_{2021}}{1+a_{2021}{a}_{2022}}\right)$
$=\left[\left({\tan }^{-1}{a}_2\right)-{\tan }^{-1}{a}_1\right]+\left[{\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2\right]+\dots ..+\left[{\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_{2021}\right]$
$={\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_1$
$={\tan }^{-1}(2022)-{\tan }^{-1}1={\tan }^{-1}2022-\frac{\pi }{4}\text{ (option 3) }$
$=\left(\frac{\pi }{2}-{\cot }^{-1}(2022)\right)-\frac{\pi }{4}$
$=\frac{\pi }{4}-{\cot }^{-1}(2022)(\text{ option 1) }$