Trigonometry : Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$ Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
 A: The same question is answered here with $\csc^2(4\pi/7)$ instead of $\csc^2(3\pi/7)$.
A: I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $\tan$ and $\cot$ need to reinforce. 
First, some preparation. 
We all know that 
$$
\tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}. 
$$
Then 
$$
\cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot (y) - \cot (x)},
$$
or
$$
\cot (x) \cot (y) = \cot(x-y) (\cot(y) -\cot(x)) - 1
$$
Especially,
$$
\cot (2x) = \frac {\cot^2 (x) -1} {2 \cot (x)},
$$
which yields
$$
\cot^2(x) - 1 = 2\cot(x)\cot(2x).
$$
Also, 
$$
\csc^2(x) = \frac {\sin^2(x) + \cos^2(x)} {\sin ^2(x)} = 1 + \cot^2(x).
$$
Now let's start. Let $x = \pi /7$.
$$
P := \csc^2(x) + \csc^2(2x) + \csc^2(4x) = 3 + \cot^2(x) + \cot^2(2x) + \cot^2(3x).
$$
By the formula above, 
$$
P = 6 + 2(\cot(x) \cot(2x) + \cot(2x) \cot (4x) + \cot (4x) \cot (8x)) =: 6 +2Q. 
$$
Now change the angle: since $\cot(\pi \pm y) = \mp\cot (y)$, we have
$$
Q = \cot(x)\cot(2x) - \cot(2x) \cot (3x) - \cot(3x) \cot (x).
$$
Now,
\begin{align*}
Q &= 1+ \cot(x)(\cot(x) - \cot(2x)) - \cot(x) (\cot(2x) -\cot(3x)) - \cot(2x)(\cot(x) - \cot (3x))\\
&=   1+  \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + \cot(2x) \cot(3x)\\
&=  \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + (\cot(2x)-\cot(3x))\cot(x) \\
&= \cot(x)(\cot(x) -2\cot(2x))\\
&= \cot^2(x) - 2\cot(2x) \cot(x)\\
&=1.
\end{align*}
Therefore $P = 6+2Q=8$. 
