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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+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\cdots .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)  is equal to

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{a}_{1}=1, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots . \mathrm{a}_{\mathrm{n}} be consecative natural numbers.
\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)
\Rightarrow \mathrm{T}_{\mathrm{K}}=\tan ^{-1}\left(\frac{1}{1+\mathrm{K}(\mathrm{K}+1)}\right)
              \quad=\tan ^{-1}\left(\frac{\mathrm{K}+1-\mathrm{K}}{1+\mathrm{K}(\mathrm{K}+1)}\right)
             \quad=\tan ^{-1}(\mathrm{~K}+1)-\tan ^{-1} \mathrm{~K}

\mathrm{~T}_{1}=\tan ^{-1} 2-\tan ^{-1} 1
\mathrm{~T}_{2}=\tan ^{-1} 3-\tan ^{-1} 2
\mathrm{~T}_{3}=\tan ^{-1} 4-\tan ^{-1} 3
\vdots
\underline{\mathrm{T}_{2021}=\tan ^{-1}(2022)-\tan ^{-1}(2021)}
On adding

\sum \mathrm{T}_{\mathrm{n}}=\tan ^{-1}(2022)-\tan ^{-1}(1)

\sum_{\mathrm{n}=1}^{2021} \mathrm{~T}_{\mathrm{n}}=\tan ^{-1}(2022)-\frac{\pi}{4}

Posted by

Pankaj Sanodiya

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