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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The value of the expression ^ -1 (1)/(2) + ^ -1 and (9)/(2) + ^ -1...

The value of the expression cot⁡−112+cot⁡−1 and 92+cot⁡−1 and 252+cot⁡−1 and 492+.......\cot^{-1 } \dfrac{1}{2} + \cot^{-1 \text{ and } } \dfrac{9}{2} + \cot^{-1 \text{ and } } \dfrac{2 5}{2} + \cot^{-1 \text{ and } } \dfrac{4 9}{2} + \text{.......} upto nn terms is

Options

  1. Atan⁡−12n\tan^{-1} 2n
  2. Btan⁡−1(2n−1)\tan^{-1} \left(2n - 1\right)
  3. Ctan⁡−1n\tan^{-1} n
  4. Dtan⁡−12n−tan⁡−11\tan^{-1} 2n - \tan^{-1} 1

Correct answer

A. tan⁡−12n\tan^{-1} 2n

Step-by-step solution

Given expression& =tan⁡−12+tan⁡−129+tan⁡−1225+tan⁡−1249+........= \tan^{-1 } 2 + \tan^{-1 } \dfrac{2}{9} + \tan^{-1 } \dfrac{2}{2 5} + \tan^{-1 }\dfrac{2}{4 9} + \text{........}

General term=2(2n−1)2=24n2−4n+1=21+4n(n−1)=2n−(2n−2)1+2n(2n−2)\text{General term} = \dfrac{2}{\left(2n - 1\right)^{2}} = \dfrac{2}{4n^{2} - 4n + 1} = \dfrac{2}{1 + 4n\left(n - 1\right)} = \dfrac{2n - \left(2n - 2\right)}{1 + 2n \left(2n - 2\right)}

Tn=tan⁡−12n−tan⁡−1(2n−2){T}_{n} = \tan^{-1} 2n - \tan^{-1} \left(2n - 2\right)

∴∴ & Sum of the series =tan⁡−12−tan⁡−10+tan⁡−14−tan⁡−12+tan⁡−16−tan⁡−14+…tan⁡−12n−tan⁡−1(2n−2)= \tan^{-1} 2 - \tan^{-1} 0 + \tan^{-1} 4 - \tan^{-1} 2 + \tan^{-1} 6 - \tan^{-1} 4 + \ldots \tan^{-1} 2n - \tan^{-1} \left(2n - 2\right)

=tan⁡−12n−tan⁡−10=tan⁡−12n= \tan^{-1} 2n - \tan^{-1} 0 = \tan^{-1} 2n

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