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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2020 · 5 September · Morning Shift · Mathematics · Inverse Trigonometric Functions

If S is the sum of the first 10 terms of the series, ^ - 1...

If SS is the sum of the first 10 terms of the series, tan⁡−1(13)+tan⁡−1(17)+tan⁡−1(113)+tan⁡−1(121)+……\tan^{- 1}\left(\dfrac{1}{3}\right) + \tan^{- 1}\left(\dfrac{1}{7}\right) + \tan^{- 1}\left(\dfrac{1}{13}\right) + \tan^{- 1}\left(\dfrac{1}{21}\right) + \ldots \ldots then tan⁡(S)\tan ( S ) is equal to :

Options

  1. A56\dfrac{5}{6}
  2. B511\dfrac{5}{11}
  3. C−56- \dfrac{5}{6}
  4. D1011\dfrac{10}{11}

Correct answer

A. 56\dfrac{5}{6}

Step-by-step solution

S=tan⁡−113+tan⁡−117+tan⁡−1113+……S = \tan^{- 1}\dfrac{1}{3} + \tan^{- 1}\dfrac{1}{7} + \tan^{- 1}\dfrac{1}{13} + \ldots \ldots up to1010 terms.

S=tan⁡−1(2−11+1.2)+tan⁡−1(3−21+2.3)+tan⁡−1(4−31+3.4)+……+tan⁡−1(11−101+11.10)S = \tan^{- 1}\left(\dfrac{2 - 1}{1 + 1 . 2}\right) + \tan^{- 1}\left(\dfrac{3 - 2}{1 + 2 . 3}\right) + \tan^{- 1}\left(\dfrac{4 - 3}{1 + 3 . 4}\right) + \ldots \ldots + \tan^{- 1}\left(\dfrac{11 - 10}{1 + 11 . 10}\right)

⇒S=(tan⁡−12−tan⁡−11)+(tan⁡−13−tan⁡−12)+……+(tan⁡−111−tan⁡−110)⇒ S = \left(\tan^{- 1}2 - \tan^{- 1}1\right) + \left(\tan^{- 1}3 - \tan^{- 1}2\right) + \ldots \ldots + \left(\tan^{- 1}11 - \tan^{- 1}10\right)

⇒S=tan⁡−111−tan⁡−11⇒ S = \tan^{- 1}11 - \tan^{- 1}1

⇒S=tan⁡−1(11−11+11⋅1)=tan⁡−1(1012)⇒ S = \tan^{- 1}\left(\dfrac{11 - 1}{1 + 11 \cdot 1}\right) = \tan^{- 1}\left(\dfrac{10}{12}\right)

∴tan⁡(S)=56∴ \tan ( S ) = \dfrac{5}{6}

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