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\bg_green \hspace{-16}$Let $\mathbf{\frac{\pi}{7}=\theta\Leftrightarrow 7\theta=\pi}$\\\\ $\mathbf{4\theta=\pi-3\theta}$\\\\ Now taking $\mathbf{\tan}$ on both side, We Get\\\\ $\mathbf{\tan(4\theta)=\tan(\pi-3\theta)=-\tan(3\theta)}$\\\\ $\mathbf{\frac{2\than(\tan (2\theta))}{1-\tan^2(2\theta)}=-\tan(3\theta)}$\\\\\\ $\mathbf{\frac{4-4\tan^2\theta}{1+\tan^4\theta-6\tan^2\theta}=\frac{\tan^2\theta-3}{1-3\tan^2\theta}}$\\\\ $\tan^6\theta-21\tan^4\theta+35\tan^2\theta-7=0$\\\\ Now Let $\mathbf{\tan^2\theta=y}\;,$ and $\mathbf{\theta=\frac{\pi}{7}\;,\frac{2\pi}{7}\;\frac{3\pi}{7}\;}$ Then \\\\ $\mathbf{y^3-21y^2+35y-7=0}$\\\\ So Let $\mathbf{y_{1}=\tan^2\left(\frac{\pi}{7}\right)\;\;,y_{2}=\tan^2\left(\frac{2\pi}{7}\right)\;\;,y_{3}=\tan^2\left(\frac{3\pi}{7}\right)\;}$ are the roots\\\\ of given equation.\\\\ So $\mathbf{y_{1}+y_{2}+y_{3}=\frac{21}{1}}$\\\\\\ So $\mathbf{\tan^2\left(\frac{\pi}{7}\right)+\tan^2\left(\frac{2\pi}{7}\right)+\tan^2\left(\frac{3\pi}{7}\right)=21}$

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