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JEE Main 2019
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Hard

Question

The value of cot(n=119cot1(1+p=1n2p))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{p = 1}^n {2p} } \right)} \right) is :

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Solution

cot(n=119cot1(1+n(n+1))\cot \left( {\sum\limits_{n = 1}^{19} {{{\cot }^{ - 1}}\left( {1 + n\left( {n + 1} \right)} \right.} } \right) cot(n=119cot1(n2+n+1))=cot(n=119tan111+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(tan1(n+1)tan1n)\sum\limits_{n = 1}^{19} {\left( {{{\tan }^{ - 1}}\left( {n + 1} \right) - {{\tan }^{ - 1}}n} \right)} cot(tan120tan11)=cotAcotβ+1cotβcotA\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)+11120=2119{{1\left( {{1 \over {20}}} \right) + 1} \over {1 - {1 \over {20}}}} = {{21} \over {19}}

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