Verified Solution • 2025–26 Session moderate

QJEEM19Ma33ed: JEE Main 2019 Mathematics - Class 12 JEEM19Ma33ed Solution: The value of  is :

The value of cot⁡(∑n=119cot⁡−1(1+∑p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right)  is :

Multiple Choice Options

  • A 2223{{22} \over {23}}
  • B 2322{{23} \over {22}}
  • C 2119{{21} \over {19}}
  • D 1921{{19} \over {21}}

Step-by-Step Verified Solution

cot⁡(∑n=119cot⁡−1(1+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right)

cot⁡(∑n=119cot⁡−1(n2+n+1))=cot⁡(∑n=119tan⁡−111+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {{n^2} + n + 1} \right)} } \right) = \cot \left( {\sum\limits_{n = 1}^{19} {{{\tan }^{ - 1}}{1 \over {1 + n\left( {n + 1} \right)}}} } \right)

∑n=119(tan⁡−1(n+1)−tan⁡−1n)\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)}

cot⁡(tan⁡−120−tan⁡−11)=cot⁡Acot⁡β+1cot⁡β−cot⁡A\cot \left( {{{\tan }^{ - 1}}20 - {{\tan }^{ - 1}}1} \right) = {{\cot A\cot \beta + 1} \over {\cot \beta - \cot A}}

(Where tanA == 20, tanB == 1)    1(120)+11−120=2119{{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring