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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Inverse Trigonometric Functions

The value of cot ( ∑ _ n = 1 ^ 19 ^ - 1 (1 + ∑ _ p = 1 ^ n 2p)) is:

The value of cotcot\left(\mathop{∑}_{n = 1}^{19}\cot^{- 1} \left(1 + \mathop{∑}_{p = 1}^{n}2p\right)\right) is:

Options

  1. A2119\dfrac{21}{19}
  2. B1921\dfrac{19}{21}
  3. C2322\dfrac{23}{22}
  4. D2223\dfrac{22}{23}

Correct answer

A. 2119\dfrac{21}{19}

Step-by-step solution

We have, cot∑n=119(cot⁡−1(1+∑p=1n2p))cot \mathop{∑}_{n = 1}^{19}\left(\cot^{- 1} \left(1 + \mathop{∑}_{p = 1}^{n}2p\right)\right)

=cot∑n=119tan⁡−1(11+n(n+1))= cot\mathop{∑}_{n = 1}^{19}\tan^{- 1}\left(\dfrac{1}{1 + n\left(n + 1\right)}\right)

=cot∑n=119tan⁡−1(n+1−n1+n(n+1))= cot\mathop{∑}_{n = 1}^{19}\tan^{- 1}\left(\dfrac{n + 1 - n}{1 + n\left(n + 1\right)}\right)

=cot∑n=119(tan⁡−1(n+1)−tan⁡−1(n))= cot \mathop{∑}_{n = 1}^{19}( \tan^{- 1}\left(n + 1\right) - \tan^{- 1}\left(n\right) )

=cot⁡(tan⁡−120−tan⁡−11)= \cot \left(\tan^{- 1}20 - \tan^{- 1}1\right)

=1tan⁡(tan⁡−120−tan⁡−11)= \dfrac{1}{\tan ( \tan^{- 1}20 - \tan^{- 1}1 )}

=120−11+(20).1=2119= \dfrac{1}{\dfrac{20 - 1}{1 + ( 20 ) . 1}} = \dfrac{21}{19}
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