Question

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

• A. $7$
• B. $\frac{35}{3}$
• C. $\frac{21}{5}$
• D. none of these

Answer 1

The correct option is C $\frac{21}{5}$

Let $7\theta =n\pi$, where $n=1,2,3$

Therefore

$\tan \theta$ can take the values $\tan \frac{\pi }{7},\tan \frac{2\pi }{7},\tan \frac{3\pi }{7}$
Now,

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

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

$\Rightarrow \frac{2\tan 2\theta }{1-{\tan }^{2}2\theta }=-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$\Rightarrow \frac{4\tan \theta (1-{\tan }^2\theta )}{{(1-{\tan }^2\theta )}^2-4{\tan }^2\theta }=-\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$\Rightarrow {\left({\tan }^2\theta \right)}^3-21{\left({\tan }^2\theta \right)}^2+35{\tan }^2\theta -7=0$
This is a cubic equation in ${\tan }^2\theta$
Sum of the roots of the equation:

${\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7}=21$
Cubic equation can also be written as

$7{\left({\cot }^2\theta \right)}^3-35{\left({\cot }^2\theta \right)}^2+21{\cot }^2\theta -1=0$

Sum of the roots:

${\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7}=\frac{35}{7}=5$
Therefore

$\frac{{\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7}}{{\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7}}=\frac{21}{5}$

Hence, option C.