Let a _ 1 = 1, a _ 2 , a _ 3 , a _ 4 , . . be consecutive natural numbers. Then ^ - 1 ( 1 1 + a _ 1 a _ 2 ) + ^ - 1 ( 1 1 + a _ 2 a _ 3 ) + . . + ^ - 1 ( 1 1 + a _ 2021 a _ 2022 ) is equal to

Correct answer: A. (π)/(4) - ^ - 1 ( 2022 ) Given, ^ - 1 ( 1 1 + a _ 1 a _ 2 ) + ^ - 1 ( 1 1 + a _ 2 a _ 3 ) . . . . . . . . . . . . . . . . ^ - 1 ( 1 1 + a _ 2021 a _ 2022 ) = ^ - 1 ( a _ 2 - a _ 1 1 + a _ 1 a _ 2 ) + ^ - 1 ( a _ 3 - a _ 2 1 + a _ 2 a _ 3 ) . . . . . . . . . . . . . . . . ^ - 1 ( a _ 2022 - a _ 2021 1 + a _ 2021 a _ 2022 ) = ^ - 1 a _ 2 - ^ - 1 a _ 1 + ^ - 1 a _ 3 - ^ - 1 a _ 2 . . . . . . . . . . + ^ - 1 a _ 2022 - ^ - 1 a _ 2021 \ using the formula ^ - 1 x - ^ - 1 y = ^ - 1 ((x - y)/(1 + xy))\ = ^ - 1 a _ 2022 - ^ - 1 a _ 1 = ^ - 1 a _ 2022 - ^ - 1 1\ as given a _ 1 = 1\ = ^ - 1 a _ 2022 - (π)/(4) = (π)/(2) - ^ - 1 a _ 2022 - (π)/(4) = (π)/(4) - ^ - 1 a _ 2022 = (π)/(4) - ^ - 1 2022 As a _ 1 = 1 , a _ 2 = 2 .... and so on a _ 2022 = 2022

JEE Main 202330 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 30 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main 2023 · 30 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

Let a _ 1 = 1, a _ 2 , a _ 3 , a _ 4 , . . be consecutive natural...

Let

{a}_{1} = 1, {a}_{2}, {a}_{3}, {a}_{4}, \ldots .

. be consecutive natural numbers. Then

\tan^{- 1}

\left(\dfrac{1}{1 + {a}_{1}{a}_{2}}\right) + \tan^{- 1}\left(\dfrac{1}{1 + {a}_{2}{a}_{3}}\right)

+ \ldots . . + \tan^{- 1}\left(\dfrac{1}{1 + {a}_{2021}{a}_{2022}}\right)

is equal to

Options

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

Correct answer

A. $\dfrac{\pi}{4} - \cot^{- 1} ( 2022 )$

Step-by-step solution

Given,

$\tan^{- 1}\left(\dfrac{1}{1 + {a}_{1}{a}_{2}}\right) + \tan^{- 1}\left(\dfrac{1}{1 + {a}_{2}{a}_{3}}\right) . . . . . . . . . . . . . . . . \tan^{- 1}\left(\dfrac{1}{1 + {a}_{2021}{a}_{2022}}\right)$

$= \tan^{- 1}\left(\dfrac{{a}_{2} - {a}_{1}}{1 + {a}_{1}{a}_{2}}\right) + \tan^{- 1}\left(\dfrac{{a}_{3} - {a}_{2}}{1 + {a}_{2}{a}_{3}}\right) . . . . . . . . . . . . . . . . \tan^{- 1}\left(\dfrac{{a}_{2022} - {a}_{2021}}{1 + {a}_{2021}{a}_{2022}}\right)$

$= \tan^{- 1}{a}_{2} - \tan^{- 1}{a}_{1} + \tan^{- 1}{a}_{3} - \tan^{- 1}{a}_{2} . . . . . . . . . . + \tan^{- 1}{a}_{2022} - \tan^{- 1}{a}_{2021}$

\left\{\text{using the formula}\tan^{- 1}x - \tan^{- 1}y = \tan^{- 1}\left(\dfrac{x - y}{1 + xy}\right)\right\} $$ = \tan^{- 1}{a}_{2022} - \tan^{- 1}{a}_{1} $$ = \tan^{- 1}{a}_{2022} - \tan^{- 1}1\left\{\text{as given}{a}_{1} = 1\right\} $$ = \tan^{- 1}{a}_{2022} - \dfrac{\pi}{4} $$ = \dfrac{\pi}{2} - \cot^{- 1}{a}_{2022} - \dfrac{\pi}{4} $$ = \dfrac{\pi}{4} - \cot^{- 1}{a}_{2022} $$ = \dfrac{\pi}{4} - \cot^{- 1}2022 $As$ {a}_{1} = 1 $,$ {a}_{2} = 2 $.... and so on$ {a}_{2022} = 2022$

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