Question

Find the sum of digits of the numerical value of $({\tan }^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7})({\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7})$.

Answer 1

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

$\therefore 4\theta +3\theta =\pi$

$\Rightarrow \tan 4\theta =\tan (\pi -30)$

$\Rightarrow \tan 4\theta =-\tan 3\theta$

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

$\Rightarrow \frac{4z-4z^3}{1-6z^2+z^4}=\frac{-3z-z^3}{1-3z^2}$ [where $\tan \theta =z$]

$\Rightarrow (4-4z^2)(1-3z^2)=-(3-z^2)(1-6z^2+z^4)$

$\Rightarrow z^6-21z^4+35z^2-7=0$----------------------------------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$=\frac{-(-21)}{1}=21$

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

or, $7y^6-35y^4+21y^2-1=0$--------------3

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

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

From equation 3, sum of roots of Equation 3$=\frac{35}{7}$

$\Rightarrow {\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7}=5$------------------------3

$\therefore By multiplying equation 2 & 4, we get

$(ta{n}^2\frac{\pi }{7}+{\tan }^2\frac{2\pi }{7}+{\tan }^2\frac{3\pi }{7})({\cot }^2\frac{\pi }{7}+{\cot }^2\frac{2\pi }{7}+{\cot }^2\frac{3\pi }{7})$

$=21\times 5=105$