Question

The value of ${\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7}$ is

• A. $21$
• B. $147$
• C. $42$
• D. none of these

Answer 1

The correct option is A $21$

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

or, $4\theta +3\theta =n\pi$

or, $\tan 4\theta =\tan (n\pi -3\theta )$

or, $\tan 4\theta =\tan 3\theta$

or, $\frac{4\tan \theta -4\tan 3\theta }{1-6{\tan }^6\theta +{\tan }^4\theta }=-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

or, $\frac{4z-4z^3}{1-6z^2+z^4}=-\frac{3z-z^3}{1-3z^2}$

or, $4(z-4z^2)(1-3z^2)=-(3-z^2)\times (1-6z^3+z^4)$

or, $z^6-21z^4+35z^2-7=0$---------------------1

This is a cubic equation in $z^2$ i.e. in ${\tan }^2\theta$.

The roots of this equation are $\therefore {\tan }^2\frac{\pi }{7},{\tan }^2\frac{2\pi }{7},{\tan }^2\frac{3\pi }{7}$

From Equation 1, sum of roots $=\frac{-(-21)}{1}=21$

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