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Question Number 181998 by Tolmasbek last updated on 03/Dec/22

	Answered by a.lgnaoui last updated on 03/Dec/22
	$α=3((π/(18))+((πn)/9)) tan α=((tan ((π/(18))+((πn)/9))[3−tan^2 ((π/(18))+((πn)/9))])/(1−3tan^2 ((π/(18))+((πn)/9)))) =((tan ((π/9)((1/2)+n)[3−tan^2 ((π/9)((1/2)+n)])/(1−3tan^2 ((π/9)((1/2)+n))) { (((π/9)((1/2)+n)≠(π/2)+kπ)),((1−3tan^2 ((π/9)((1/2)+n))≠0)) :} { (( n ≠4+9k (k∈Z^+ ))),((n≠1+k and n≠9k−2 )) :} donc n≠2k k∈Z k>1 So, α is not also true$
	$α = 3 (\frac{π}{18} + \frac{π n}{9})$ $\tan α = \frac{\tan (\frac{π}{18} + \frac{π n}{9}) [3 - \tan^{2} (\frac{π}{18} + \frac{π n}{9})]}{1 - 3 \tan^{2} (\frac{π}{18} + \frac{π n}{9})}$ $= \frac{\tan (\frac{π}{9} (\frac{1}{2} + n) [3 - \tan^{2} (\frac{π}{9} (\frac{1}{2} + n)]}{1 - 3 \tan^{2} (\frac{π}{9} (\frac{1}{2} + n)}$ ${\begin{cases} \frac{π}{9} (\frac{1}{2} + n) \neq \frac{π}{2} + k π \\ 1 - 3 \tan^{2} (\frac{π}{9} (\frac{1}{2} + n)) \neq 0 \end{cases}$ ${\begin{cases} n \neq 4 + 9 k (k \in Z^{+}) \\ n \neq 1 + k a n d n \neq 9 k - 2 \end{cases}$ $d o n c n \neq 2 k k \in Z k > 1$ $S o, α i s n o t a l s o t r u e$
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