Let a1 = 1, a2, a3, a4, ........ be consecutive natural numbers.Then tan−1⁡11+a1a2+tan−1⁡11+a2a3+…..+tan−1⁡11+a2021a2022 is equal to

tan−1⁡a2−a11+a1a2+…..+tan−1⁡a2022−a20211+a2021a2022=tan−1⁡a2−tan−1⁡a1+…..+tan−1⁡a2022−tan−1⁡a2021=tan−1⁡a2022−tan−1⁡a1=tan−1⁡a2022−π4 =π2−cot−1⁡a2022−π4∵a1=1 =π4−cot−1⁡a2022=π4−cot−1⁡2022 Ans: A,C

Let a1 = 1, a2, a3, a4, ........ be consecutive natural numbers.Then tan−1⁡11+a1a2+tan−1⁡11+a2a3+…..+tan−1⁡11+a2021a2022 is equal to

tan−1⁡a2−a11+a1a2+…..+tan−1⁡a2022−a20211+a2021a2022=tan−1⁡a2−tan−1⁡a1+…..+tan−1⁡a2022−tan−1⁡a2021=tan−1⁡a2022−tan−1⁡a1=tan−1⁡a2022−π4 =π2−cot−1⁡a2022−π4∵a1=1 =π4−cot−1⁡a2022=π4−cot−1⁡2022 Ans: A,C

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Let a₁ = 1, a₂, a₃, a₄, ........ 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

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$\tan^{- 1} (2022) - \frac{π}{4}$

$\frac{π}{4} - \tan^{- 1} (2022)$

$\cot^{- 1} (2022) - \frac{π}{4}$

$\frac{π}{4} - \cot^{- 1} (2022)$

answer is A.

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Detailed Solution

$\begin{array}{l} \tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \dots . . + \tan^{- 1} (\frac{a_{2022} - a_{2021}}{1 + a_{2021} a_{2022}}) \\ = \tan^{- 1} a_{2} - \tan^{- 1} a_{1} + \dots . . + \tan^{- 1} (a_{2022}) - \tan^{- 1} (a_{2021}) \\ = \tan^{- 1} (a_{2022}) - \tan^{- 1} a_{1} = \tan^{- 1} (a_{2022}) - \frac{π}{4} = \frac{π}{2} - \cot^{- 1} (a_{2022}) - \frac{π}{4} (∵ a_{1} = 1) = \frac{π}{4} - \cot^{- 1} (a_{2022}) = \frac{π}{4} - \cot^{- 1} (2022) A n s : A, C \end{array}$