2021/070) Evaluate $\dfrac{1}{1-\cos \dfrac{\pi}{9}}+\dfrac{1}{1-\cos \dfrac{5\pi}{9}}+\dfrac{1}{1-\cos \dfrac{7\pi}{9}}$.

We know that $\cos \frac{\pi}{9},\cos \frac{5\pi}{9},\cos \frac{7\pi}{9}$ are different and they are

roots of equation $\cos 3x = \cos \frac{\pi}{3} = \frac{1}{2}$

or $4\cos^3 x - 3\cos\,x =\frac{1}{2}$

or

so $\cos\frac{\pi}{9}, \cos\frac{5\pi}{9}, \cos\frac{7\pi}{9}$ are roots of equation

$x^3 - \frac{3}{4}x - \frac{1}{8}= 0$

let $x_1= \cos\frac{\pi}{9}, x_2 = \cos\frac{5\pi}{9}, x_3=\cos\frac{7\pi}{9}$

Now $x_1,x_2,x_3$ are roots of equation

$f(x) = x^3 - \frac{3}{4}x - \frac{1}{8}= 0\cdots(1)$

by Vieta's formula we have

$x_1 + x_2 + x_3 = 0\cdots(2)$

$x_1 x_2 + x_2x_3 + x_3 x_1 = \frac{-3}{4}\cdots(3)$

Further $f(1) = (1-x_1)(1-x_2)(1-x_3) =  1- \frac{1}{4} - \frac{1}{8} = \frac{1}{8}\cdots(3)$

And we need to evaluate $\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

Now

$\frac{1}{1-x_1 } + \frac{1}{1-x_2} + \frac{1}{1-x_3}$

$= \frac{(1-x_2)(1-x_3) + (1-x_1)(1-x_3) + (1-x_1)(1-x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{1-x_2 - x_3 + x_2x_3 + 1-x_1 - x_3  + x_1x_3  + 1-x_1 - x_2 + x_1x_2}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2(x_1 + x_2 + x_3)  + (x_2x_3 + x_3x_1  + x_1x_2)}{(1-x_1)(1-x_2)(1-x_3)}$

$= \frac{3 - 2 * 0 + \frac{-3}{4}}{\frac{1}{8}}$ putting the values using (2) , (3) and (4)

$= 18$

Hence $\frac{1}{1-\cos \frac{\pi}{9}} + \frac{1}{1-\cos \frac{5\pi}{9}} + \frac{1}{1-\frac{7\pi}{9}}= 18$