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If $ω (\neq 1)$ is a cube root of unity, then the value of $\tan [(ω^{2017} + ω^{2225}) π - π / 3]$

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$- \sqrt{3}$

$- \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

answer is C.

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Detailed Solution

$ω^{2017} = {(ω^{3})}^{672} ω = ω$

and $ω^{2225} = {(ω^{3})}^{741} ω^{2} = ω^{2}$

$∴ \tan [(ω^{2017} + ω^{2225}) π - π / 3]$

$= \tan [(ω + ω^{2}) π - π / 3]$

$= \tan (- π - π / 3) = - \tan (π + π / 3)$

$= - \tan (π / 3) = - \sqrt{3}$

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