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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Advanced 2013 · Paper 1 · Mathematics · Inverse Trigonometric Functions

The value of cot ( Σ _ n = 1 ^ 23 cot ^ - 1 (1 + Σ _ k = 1 ^ n 2k))...

The value of cotcot\left(\mathop{Σ}_{\text{n} = 1}^{23}{\text{cot}}^{- 1}\left(1 + \mathop{Σ}_{\text{k} = 1}^{\text{n}}2\text{k}\right)\right) is

Options

  1. A2325\dfrac{2 3}{2 5}
  2. B2523\dfrac{2 5}{2 3}
  3. C2324\dfrac{2 3}{2 4}
  4. D2423\dfrac{2 4}{2 3}

Correct answer

B. 2523\dfrac{2 5}{2 3}

Step-by-step solution

cot⁡(Σn=123cot−1(1+Σk=1n2k))=cotΣn=123cot−1(1+2+4+6+…+2n)\cot \left(\mathop{Σ}_{\text{n} = 1}^{23}{\text{cot}}^{- 1}\left(1 + \mathop{Σ}_{\text{k} = 1}^{\text{n}}2\text{k}\right)\right) = \text{cot}\mathop{Σ}_{\text{n} = 1}^{23}{\text{cot}}^{- 1}\left(1 + 2 + 4 + 6 + \ldots + 2\text{n}\right)
\text{=cot}Σ{\text{cot}}^{- 1}\left(1 + \text{n}\left(\text{n} + 1\right)\right)<br/><br />\text{=cot}Σ{\text\tan}^{- 1}\dfrac{\left(\text{n} + 1\right) - \text{n}}{1 + \text{n}\left(\text{n} + 1\right)}(cot−1x=tan⁡−11x)\left(co{t}^{- 1}x = \tan^{- 1}\dfrac{1}{x}\right)
\text{=cot}\mathop{Σ}_{\text{n} = 1}^{23}\left({\text\tan}^{- 1}\left(\text{n} + 1\right) - {\text\tan}^{- 1}\text{n}\right)\left[{\text\tan}^{\text{-1}}\left(\dfrac{\text{x-y}}{1 + xy}\right){\text{=tan}}^{\text{-1}}{\text{x-tan}}^{\text{-1}}\text{y}\right]

Now {\text\tan}^{- 1}\left(\text{n} + 1\right) - {\text\tan}^{- 1}{\text{n=tan}}^{- 1}\left(2\right) - {\text\tan}^{- 1}{\text{1+tan}}^{- 1}\left(3\right) - {\text\tan}^{- 1}{\text{2+....+tan}}^{- 1}\left(24\right) - {\text\tan}^{- 1}\text{23}

\text{=cot}\left({\text\tan}^{- 1}24 - {\text\tan}^{- 1}1\right)
\text{=cot}\left({\text\tan}^{- 1}\dfrac{24 - 1}{1 + 24}\right)
=cot(cot−12523)=2523\text{=cot}\left({\text{cot}}^{- 1}\dfrac{25}{23}\right) = \dfrac{25}{23}

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