If tan left(frac{pi}{9}right), x, tan left(fr | vedclass

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(A) Given that $	an \left(\frac{\pi}{9}ight), x, 	an \left(\frac{7 \pi}{18}ight)$ are in arithmetic progression,we have $2x = 	an \left(\frac{\

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(A) Given that $	an \left(\frac{\pi}{9}ight), x, 	an \left(\frac{7 \pi}{18}ight)$ are in arithmetic progression,we have $2x = 	an \left(\frac{\pi}{9}ight) + 	an \left(\frac{7 \pi}{18}ight)$,so $x = \frac{1}{2} \left(	an \frac{\pi}{9} + 	an \frac{7 \pi}{18}ight)$.Given that $	an \left(\frac{\pi}{9}ight), y, 	an \left(\frac{5 \pi}{18}ight)$ are in arithmetic progression,we have $2y = 	an \left(\frac{\pi}{9}ight) + 	an \left(\frac{5 \pi}{18}ight)$.We need to find $|x - 2y|$.Substituting the expressions:$x - 2y = \frac{1}{2} \left(	an \frac{\pi}{9} + 	an \frac{7 \pi}{18}ight) - \left(	an \frac{\pi}{9} + 	an \frac{5 \pi}{18}ight)$.Using the identities $	an \left(\frac{7 \pi}{18}ight) = \cot \left(\frac{\pi}{9}ight)$ and $	an \left(\frac{5 \pi}{18}ight) = \cot \left(\frac{2 \pi}{9}ight)$:$x - 2y = \frac{1}{2} 	an \frac{\pi}{9} + \frac{1}{2} \cot \frac{\pi}{9} - 	an \frac{\pi}{9} - \cot \frac{2 \pi}{9} = \frac{1}{2} \left(\cot \frac{\pi}{9} - 	an \frac{\pi}{9}ight) - \cot \frac{2 \pi}{9}$.Since $\cot 	heta - 	an 	heta = 2 \cot(2	heta)$:$x - 2y = \frac{1}{2} \left(2 \cot \frac{2 \pi}{9}ight) - \cot \frac{2 \pi}{9} = \cot \frac{2 \pi}{9} - \cot \frac{2 \pi}{9} = 0$.Therefore,$|x - 2y| = 0$.

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

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