2020/021) Evaluate $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}$

Let $\theta = \frac{\pi}{7}$

We need to find $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta$ 
Now $7\theta = \pi$

Or $4\theta = \pi - 3\theta$

Taking $\tan$ of both sides we get

Or $\tan 4\theta = \tan (\pi - 3\theta) = - \tan 3\theta$

Or $\frac{4\tan \theta - 4\tan^3 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} - \frac{3\tan \theta - \tan ^3\theta}{1-3\tan ^2 \theta}$

Or  $\frac{4 - 4\tan^2 \theta}{1-6\tan ^2 \theta + \tan ^4 \theta} + \frac{3 - \tan ^2\theta}{1-3\tan ^2 \theta}$

 Or  $(4 - 4\tan^2 \theta)(1-3\tan ^2 \theta) + (1-6\tan ^2 \theta + \tan ^4 \theta)(3 - \tan ^2\theta)$

Or $4- 16 \tan ^2 \theta + 12 \tan ^4 \theta + 3  -19 \tan ^2 \theta + 9\tan^4 \theta - \tan ^26\theta = 0$

Or $\tan ^6\theta  - 21\tan ^4 \theta + 35 \tan ^2 \theta -7=0$

The above equation is a cubic equation in $\tan^2 \theta$ whose roots are $\tan \theta$, $\tan 2\theta$, and 

$\tan 3\theta$,

so using Vieta's formula $\tan^2 \theta + \tan^2 2\theta + \tan^2 3\theta= 21$ 

or $\tan^2 \frac{\pi}{7} + \tan^2 \frac{2\pi}{7} + \tan^2 \frac{3\pi}{7}= 21$ 

as a corollary adding 1 to each term of LHS and so adding 3 to RHS we get

 $\sec^2 \frac{\pi}{7}+ \sec^2 \frac{2\pi}{7} + \sec^2 \frac{3\pi}{7}= 24$