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JEE Main 2018
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Hard

Question

The value of cot(n=150tan1(11+n+n2))\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right) is :

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Solution

cot(n=150tan1(11+n+n2))\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right) =cot(n=150tan1((n+1)n1+(n+1)n)) = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right) =cot(n=150(tan1(n+1)tan1n) = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right) =cot(tan151tan11) = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1) =cot(tan1(5111+51)) = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right) =cot(cot1(5250)) = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right) =2625 = {{26} \over {25}}

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