Ifn tan( (pi / 9) ),x,tan( (7pi / 18) )n are in arithmetic progression andn tan( (pi / 9) ),y,tan( (5pi / 18) )n are also in arithmetic progression, then |x - 2y| is equal to

(d) : We have, tan ((pi / 9)), x, tan ((7 pi / 18)) in A.P.nn n => 2 x =tan ((pi / 9))+tan ((7 pi / 18)) n => 2 x=tan ((pi / 9))+tan [(pi / 2)-(pi / 9)] => x=(1 / 2)(tan (pi / 9)+cot (pi / 9)) \ (i)n nnnJEE Simitarly y=(1 / 2)[tan (pi / 9)+tan (5 pi / 18)] n => 2 y=[tan (pi / 9)+tan ((pi / 2)-(2 pi / 9))] => 2 y=tan (pi / 9)+cot (2 pi / 9) n Therefore |x-2 y|=|(1 / 2) tan (pi / 9)+(1 / 2) cot (pi / 9)-tan (pi / 9)-cot (2 pi / 9)| n =|(1 / 2)(cot (pi / 9)-tan (pi / 9))-cot (2 pi / 9)| n =|(1 / 2)[(1-tan ^2 pi / 9 / tan pi / 9)]-cot (2 pi / 9)|=|cot (2 pi / 9)-cot (2 pi / 9)|=0n n

If tan( \frac{\pi}{9} ),x,tan( \frac{7\pi}{18} ) ar... | Sequence & Series (Mathematics)

(d) : We have,

\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right) \in

A.P.\n

\n\begin{aligned}\n\Rightarrow 2 x & =\tan \left(\frac{\pi}{9}\right)+\tan \left(\frac{7 \pi}{18}\right) \\\n& \Rightarrow 2 x=\tan \left(\frac{\pi}{9}\right)+\tan \left[\frac{\pi}{2}-\frac{\pi}{9}\right] \Rightarrow x=\frac{1}{2}\left(\tan \frac{\pi}{9}+\cot \frac{\pi}{9}\right) \#(i)\n\end{aligned}\n

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\n\begin{aligned}\n& \text { Simitarly } y=\frac{1}{2}\left[\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}\right] \\\n& \Rightarrow 2 y=\left[\tan \frac{\pi}{9}+\tan \left(\frac{\pi}{2}-\frac{2 \pi}{9}\right)\right] \Rightarrow 2 y=\tan \frac{\pi}{9}+\cot \frac{2 \pi}{9} \\\n& \therefore|x-2 y|=\left|\frac{1}{2} \tan \frac{\pi}{9}+\frac{1}{2} \cot \frac{\pi}{9}-\tan \frac{\pi}{9}-\cot \frac{2 \pi}{9}\right| \\\n& =\left|\frac{1}{2}\left(\cot \frac{\pi}{9}-\tan \frac{\pi}{9}\right)-\cot \frac{2 \pi}{9}\right| \\\n& =\left|\frac{1}{2}\left[\frac{1-\tan ^2 \pi / 9}{\tan \pi / 9}\right]-\cot \frac{2 \pi}{9}\right|=\left|\cot \frac{2 \pi}{9}-\cot \frac{2 \pi}{9}\right|=0\n\end{aligned}\n

Explanation

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