Mathematics · 2019
JEE Main · 10 January 2019, Shift 2 · Q69
The value of cot (Σ_n =1^19 cot^-1(1+Σ_p =1^n 2 p )) is:
The value of $\displaystyle \cot \left(\sum_{\mathrm{n}=1}^{19} \cot ^{-1}\left(1+\sum_{\mathrm{p}=1}^{\mathrm{n}} 2 \mathrm{p}\right)\right)$
is :
Official answer
(3)
$\displaystyle \frac{21}{19}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.