1. Tardigrade
  2. Question
  3. Mathematics
  4. Let a1=1, a2, a3, a4, ldots .. be consecutive natural numbers. Then tan -1((1/1+a1 a2))+ tan -1((1/1+a2 a3))+ ldots .+ tan -1((1/1+a2021 a2022)) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

JEE MainJEE Main 2023Inverse Trigonometric Functions

Solution:

$a _2- a _1= a _3- a _2=\ldots . .= a _{2022}- a _{2021}=1$.
$ \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)+\ldots . .+\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]+\ldots . . +\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) }$