Mathematics - Sequence & Series Question with Solution | TestHub

$x=\frac{1}{2}\left(\tan \frac{\pi }{9}+\tan \frac{7\pi }{18}\right)$

and $2y=\tan \frac{\pi }{9}+\tan \frac{5\pi }{18}$

so, $x-2y=\frac{1}{2}\left(\tan \frac{\pi }{9}+\tan \frac{7\pi }{18}\right)$$-\left(\tan \frac{\pi }{9}+\tan \frac{5\pi }{18}\right)$

$\Rightarrow \left|x-2y\right|=\left|\frac{\cot \frac{\pi }{9}-\tan \frac{\pi }{9}}{2}-\tan \frac{5\pi }{18}\right|$

$\Rightarrow \left|x-2y\right|=\left|\frac{{\cot }^2\frac{\pi }{9}-1}{2cot\frac{\pi }{9}}-\tan \frac{5\pi }{18}\right|$

$\Rightarrow \left|x-2y\right|=\left|\cot \frac{2\pi }{9}-\tan \frac{5\pi }{18}\right|$

$=0$

$\left(\because \tan \frac{5\pi }{18}=cot\left(\frac{\pi }{2}-\frac{5\pi }{18}\right)=\cot \frac{2\pi }{9}\right)$$\left(\because \tan \frac{7\pi }{18}=cot\left(\frac{\pi }{2}-\frac{7\pi }{18}\right)=\cot \frac{\pi }{9}\right)$

$\left(\because \frac{{\cot }^2\theta -1}{2cot\theta }=cot2\theta \right)$