Mathematics · 2020
JEE Main · 2 September 2020, Shift 1 · Q53
The value of ((1+ sin (2 π)/9+i cos (2 π)/9)/(1+ sin (2 π)/9-i cos (2 π)/9))^3 is:
The value of $\displaystyle \left(\frac{1+\sin \dfrac{2 \pi}{9}+i \cos \dfrac{2 \pi}{9}}{1+\sin \dfrac{2 \pi}{9}-i \cos \dfrac{2 \pi}{9}}\right)^3$ is :
Official answer
(4)
$\displaystyle -\frac{1}{2}(\sqrt{3}-i)$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.