QJEEM23Macbb5: JEE Main 2023 Mathematics - Class 12 JEEM23Macbb5 Solution: Let . be consecutive natural numbers. Then is equal to :

Question

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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 :

Accepted Answer

a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .

. be consecutive natural numbers.

\begin{aligned}&amp; \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\\\&amp; \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \\\\&amp; \tan ^{-1}\left[\frac{1}{1+a_2 a_3}\right]=\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]=\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right)\end{aligned}

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\begin{aligned}\tan ^{-1}\left[\frac{1}{1+a_{2021} a_{2022}}\right] &amp; =\tan ^{-1}\left[\frac{1}{1+2021 \cdot 2022}\right] \\\\&amp; =\tan ^{-1}\left(\frac{1}{2021}\right)-\tan ^{-1}\left(\frac{1}{2022}\right)\end{aligned}

\therefore

\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)

=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2022}\right)=\frac{\pi}{4}-\cot ^{-1}(2022)

=\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan ^{-1}(2022)\right)=\tan ^{-1}(2022)-\frac{\pi}{4}

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