# S n n 1 2 1 tan … 3 3 – æ ö ç ÷ + è ø n n + + 1 1 tan 1 ( 19)( 20) – æ ö + ç ÷ è ø ++ + n n , then tan S is equal to :

Question:

$S=\tan ^{-1}\left(\frac{1}{n^{2}+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^{2}+3 n+3}\right)+\ldots$

$+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$, then tan $S$ is equal to :

1. $\frac{20}{401+20 n}$

2. $\frac{n}{n^{2}+20 n+1}$

3. $\frac{20}{n^{2}+20 n+1}$

4. $\frac{n}{401+20 n}$

Correct Option: 3

JEE Main Previous Year 1 Question of JEE Main from Mathematics Inverse Trigonometric Functions chapter.
JEE Main Previous Year Online April 23, 2013

Solution:

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