Question

Let $a_1=1, a_2, a_3, a_4, \ldots $ be consecutive natural numbersThen $\tan ^{-1}\left(\frac{1}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{1}{1+a_2 a_3}\right)+\ldots +\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ 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

The Correct Option is B

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