Let $\mathrm{a}_{1}=1, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}, \ldots$. be consecutive natural numbers. Then $\tan ^{-1}\left(\frac{1}{1+\mathrm{a}_{1} \mathrm{a}_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+\mathrm{a}_{2} \mathrm{a}_{3}}\right) +\ldots . .+\tan ^{-1}\left(\frac{1}{1+\mathrm{a}_{2021} \mathrm{a}_{2022}}\right)$ is equal to

(A) $\frac{\pi}{4}-\cot ^{-1}(2022)$; (C) $\tan ^{-1}(2022)-\frac{\pi}{4}$

(B) $\cot ^{-1}(2022)-\frac{\pi}{4}$

(D) $\frac{\pi}{4}-\tan ^{-1}(2022)$

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Inverse Trigonometric Functions

Let

a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots

. be consecutive natural numbers. Then

tan^{- 1} (\frac{1}{1 + a _{1} a _{2}}) + tan^{- 1} (\frac{1}{1 + a _{2} a _{3}}) + \dots .. + tan^{- 1} (\frac{1}{1 + a _{2021} a _{2022}})

is equal to

Answer: (A), (C)

As the terms are consecutive natural numbers,

a_{n + 1} - a_{n} = 1

, so

tan^{- 1} \frac{1}{1 + a _{n} a _{n + 1}} = tan^{- 1} a_{n + 1} - tan^{- 1} a_{n}

. The sum telescopes to

tan^{- 1} (a_{2022}) - tan^{- 1} (a_{1}) = tan^{- 1} (2022) - \frac{π}{4}

. Using

tan^{- 1} (2022) = \frac{π}{2} - cot^{- 1} (2022)

, this same value also equals

\frac{π}{4} - cot^{- 1} (2022)

. The sum is therefore

tan^{- 1} (2022) - \frac{π}{4}

, equivalently

\frac{π}{4} - cot^{- 1} (2022)

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · One or more correct options