Not just Trigonometry

We know that...

$\csc^2 x = 1 + \cot^2 x$

Thus, the given expression becomes:

$3 + \cot^2\frac{\pi}{7} + \cot^2\frac{2\pi}{7} + \cot^2\frac{3\pi}{7}$

Or, $3 + \sum_{r=1}^{n}\cot ^2 \frac{r\pi}{2n+1}$

Here $n=3$

Now let us assume, $\sin (2n+1)x = 0$

$\therefore (2n+1)x = m\pi$

$\therefore x = \frac{m\pi}{(2n+1)}$

Also, let us have $z= \cos x + i\sin x$

$\therefore {z}^{(2n+1)} = \cos(2n+1)x + i\sin(2n+1)x \cdots \left (1 \right )$

By, using binomial expansion, we get:-

$z^{(2n+1)} = C_0 \cos^{(2n+1)} x + C_1 (\cos^{2{n}}x).i.(\sin x) - C_2 (\cos^{2{n-1}}x)(\sin^2 x) - C_3 (\cos^{2{(n-1)}}x).i.(\sin^3 x) \cdots \left ( 2 \right )$

Equating imaginary parts of 1 and 2, we get,

$\sin(2n+1)x = C_1 (\cos^{2{n}}x)(\sin x) - C_3 (\cos^{2{(n-1)}}x)(\sin^3 x) \cdots$

Multiplying and Dividing each term of RHS to get $\cot^2 x$ , also let us assume $cot^2 x = y$

$\sin(2n+1)x = C_1 (y^n)(\sin^{(2n+1)} x) - C_3 (y^{n-1})(\sin^{2n+1} x) \cdots$

$\sin(2n+1)x = ( \sin^{2n+1}x) [C_1 (y^n) - C_3 (y^{n-1}) \cdots]$

Now we must see that in RHS $sin^{2n+1}x \neq 0$

Therefore, the second expression is zero and for finding the above sigma, we see that we need the sum of roots of the second expression amounts to the sigma operation.

Thus, sum of roots:

$\frac{-b}{a} = \frac{^{2n+1}\textrm{C}_{3}}{^{2n+1}\textrm{C}_{1}}$

$\frac{-b}{a} = \frac{(2n)(2n-1)}{6}$

Plugging in $n=3$ , we get:

$\sum_{r=1}^{n}\cot ^2 \frac{r\pi}{2n+1} = 5$

Therefore the value of the expression is equal to $3 +5 = 8$