JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
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
- A \(\frac{\pi}{4}-\cot ^{-1}(2022)\)
- B \(\cot ^{-1}(2022)-\frac{\pi}{4}\)
- C \(\tan ^{-1}(2022)-\frac{\pi}{4}\)
- D \(\frac{\pi}{4}-\tan ^{-1}(2022)\)
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}(2022)-\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\quad a _2- a _1= a _3- a _2=\ldots . .= a _{2022}- a _{2021}=1\).…
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