JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

Given,

${\tan }^{-1}\left(\frac{1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{1}{1+a_2{a}_3}\right)................{\tan }^{-1}\left(\frac{1}{1+a_{2021}{a}_{2022}}\right)$

$={\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)................{\tan }^{-1}\left(\frac{a_{2022}-a_{2021}}{1+a_{2021}{a}_{2022}}\right)$

$={\tan }^{-1}{a}_2-{\tan }^{-1}{a}_1+{\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2..........+{\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_{2021}$

$\left\{\text{using the formula } {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)\right\}$

$={\tan }^{-1}{a}_{2022}-{\tan }^{-1}{a}_1$

$={\tan }^{-1}{a}_{2022}-{\tan }^{-1}1 \left\{\text{as given} a_1=1\right\}$

$={\tan }^{-1}{a}_{2022}-\frac{\mathrm{\pi }}{4}$

$=\frac{\mathrm{\pi }}{2}-{\cot }^{-1}{a}_{2022}-\frac{\mathrm{\pi }}{4}$

$=\frac{\mathrm{\pi }}{4}-{\cot }^{-1}{a}_{2022}$

$=\frac{\mathrm{\pi }}{4}-{\cot }^{-1}2022$

As $a_1=1$, $a_2=2$ .... and so on $a_{2022}=2022$