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JEE Main 201323 AprilOnline
Exam
JEE Main
Year
2013
Date
23 April
Mode
Online
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2013 · 23 April · Online · Mathematics · Inverse Trigonometric Functions

S= ^ -1 ((1)/(n^2+n+1))+ ^ -1 ((1)/(n^2+3 n+3))+ + ^ -1...

S=tan⁡−1S=\tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}(1n2+3n+3)+…\left(\frac{1}{n^2+3 n+3}\right)+\ldots +tan⁡−1(11+(n+19)(n+20))+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right), then tan⁡S\tan S is equal to :

Options

  1. A20401+20n\frac{20}{401+20 n}
  2. Bnn2+20n+1\frac{n}{n^2+20 n+1}
  3. C20n2+20n+1\frac{20}{n^2+20 n+1}
  4. Dn401+20n\frac{n}{401+20 n}

Correct answer

C. 20n2+20n+1\frac{20}{n^2+20 n+1}

Step-by-step solution

We know that, tan⁡−111+2+tan⁡−111+2×3+tan⁡−111+3×4+……+tan⁡−111+(n−1)n+tan⁡−111+n(n+1)+……+tan⁡−111+(n+19)(n+20)=tan⁡−1n+19n+21⇒tan⁡−1n−1n+1+tan⁡−111+n(n+1)+tan⁡−111+(n+1)(n+2)+……+11+(n+19)(n+20)=tan⁡−1n+19n+21tan⁡−111+n(n+1)+tan⁡−111+(n+1)(n+2)+……+11+(n+19)(n+20)=tan⁡−1n+19n+21−tan⁡−1n−1n+1\begin{array}{r} \tan ^{-1}\frac{1}{1+2}+\tan ^{-1} \frac{1}{1+2 \times 3}+\tan ^{-1} \frac{1}{1+3 \times 4}+\ldots \ldots+\\ \tan ^{-1}\frac{1}{1+(n-1) n}+\tan ^{-1} \frac{1}{1+n(n+1)}+\ldots \ldots+\\ \tan ^{-1}\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}\\ \Rightarrow \tan ^{-1}\frac{n-1}{n+1}+\tan ^{-1} \frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}\\ +\ldots \ldots+\frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}\\ \tan ^{-1}\frac{1}{1+n(n+1)}+\tan ^{-1} \frac{1}{1+(n+1)(n+2)}+\ldots \ldots+\\ \frac{1}{1+(n+19)(n+20)}=\tan ^{-1} \frac{n+19}{n+21}-\tan ^{-1} \frac{n-1}{n+1} \end{array} tan⁡−1(1n2+n+1)+tan⁡−1(1n2+3n+3)+……+=tan⁡−1(n+19n+21−n−1n+11+n+19n+21×n−1n+1)=tan⁡−111+(n+19)(n+20)20n2+20n+1=S∴tan⁡−1 S=20n2+20n+1\begin{aligned}& \tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots \ldots+\\ & \quad=\tan ^{-1}\left(\frac{\frac{n+19}{n+21}-\frac{n-1}{n+1}}{1+\frac{n+19}{n+21} \times \frac{n-1}{n+1}}\right)\\ & \quad=\tan ^{-1} \frac{1}{1+(n+19)(n+20)} \frac{20}{n^2+20 n+1}=\mathrm{S}\\ & \therefore \quad \tan ^{-1} \mathrm{~S}=\frac{20}{n^2+20 n+1} \end{aligned}
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