If tan(π9), x, tan(7π18) , are in arthimetic progression and tan(π9), y, tan(5π18) are also in arthimetic progression, then |x−2y| is equal to:

(C) Since, tan(π9),X,tan(7π18) are in A.P.∴x=12(tanπ9+tan7π18)And tan(π9),y,tan(5π18) are in A.P.∴2y=tanπ9+tan5π18Now x−2y=12(tanπ9+tan7π18)−(tanπ5+tan5π18)⇒|x−2y|=|cotπ9−tanπ92−tan5π18|=|cot2π9−cot2π9|=0(∵tan5π18=cot2π9;tan7π18=cotπ9)

If tan(π9), x, tan(7π18) , are in arthimetic progression and tan(π9), y, tan(5π18) are also in arthimetic progression, then |x−2y| is equal to:

(C) Since, tan(π9),X,tan(7π18) are in A.P.∴x=12(tanπ9+tan7π18)And tan(π9),y,tan(5π18) are in A.P.∴2y=tanπ9+tan5π18Now x−2y=12(tanπ9+tan7π18)−(tanπ5+tan5π18)⇒|x−2y|=|cotπ9−tanπ92−tan5π18|=|cot2π9−cot2π9|=0(∵tan5π18=cot2π9;tan7π18=cotπ9)

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If $\tan (\frac{π}{9}), x, \tan (\frac{7 π}{18})$ , are in arthimetic progression and $\tan (\frac{π}{9}), y, \tan (\frac{5 π}{18})$ are also in arthimetic progression, then $| x - 2 y |$ is equal to:

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answer is C.

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Detailed Solution

(C) Since, $\tan (\frac{π}{9}), X, \tan (\frac{7 π}{18}) a r e i n A . P .$

$\begin{array}{l} ∴ x = \frac{1}{2} (\tan \frac{π}{9} + \tan \frac{7 π}{18}) \\ A n d \tan (\frac{π}{9}), y, \tan (\frac{5 π}{18}) a r e i n A . P . \\ ∴ 2 y = \tan \frac{π}{9} + \tan \frac{5 π}{18} \\ N o w x - 2 y = \frac{1}{2} (\tan \frac{π}{9} + \tan \frac{7 π}{18}) - (\tan \frac{π}{5} + \tan \frac{5 π}{18}) \\ \Rightarrow | x - 2 y | = | \frac{\cot \frac{π}{9} - \tan \frac{π}{9}}{2} - \tan \frac{5 π}{18} | \\ = | \cot \frac{2 π}{9} - \cot \frac{2 π}{9} | = 0 (∵ \tan \frac{5 π}{18} = \cot \frac{2 π}{9}; \tan \frac{7 π}{18} = \cot \frac{π}{9}) \end{array}$