JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

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) \text { is equal to }

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the given sum telescopes to

\tan^{-1}(a_{2022}) - \tan^{-1}(a_1) = \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{{\pi}}{{4}}

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)