JEE MAIN - Mathematics (2022 - 27th June Evening Shift - No. 13)
The value of $$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$ is :
$${{26} \over {25}}$$
$${{25} \over {26}}$$
$${{50} \over {51}}$$
$${{52} \over {51}}$$
Explanation
$$\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$$
$$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$$
$$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)$$
$$ = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)$$
$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)$$
$$ = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)$$
$$ = {{26} \over {25}}$$
Comments (0)
