Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that:
$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$
This is actually problem B $4371$ given at this link. Looks like a very interesting problem. 
My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a $\text{complete}$ solution.


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*First, I wrote $\sin\frac{5\pi}{14}$ as $\sin\frac{9 \pi}{14}$ so that if I put $A = \frac{\pi}{14}$ so that the given equation becomes, $$\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24$$ Then I tried working with this by taking $\text{lcm}$ and multiplying and doing something, which appeared futile.

*Next, I actually didn't work it out, but I think we have to look for a equation which has roots as $\sin$ and then use $\text{sum of roots}$ formulas to get $24$. I think I haven't explained this clearly.

  
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*$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ One post related to this is: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ I don't know how this will help as I haven't studied anything yet regarding Gauss sums.
  


This may be a $3$ year old question, but I would like to add to the list an answer that relies on the sum of $\tan^2$ identity.
Let $$\begin{align}S &= \frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}}\\
&=\frac{1}{\cos^{2}\frac{3\pi}{7}} + \frac{1}{\sin^{2}\frac{2\pi}{7}} + \frac{1}{\sin^{2}\frac{\pi}{7}}\\
&= \sec^2{\frac{3\pi}{7}} + \sec^2{\frac{2\pi}{7}} + \sec^2{\frac{\pi}{7}}\\
&= \tan^2\frac{\pi}{7} + \tan^2\frac{2\pi}{7} + \tan^2\frac{3\pi}{7} + 3\\
&= \sum_{k = 1}^3 \tan^2\frac{k\pi}{7} + 3\end{align}$$
From the sum of $\tan^2$ identity discussed over here, we have
$$
\sum_{k=1}^n\tan^2\frac{k\pi}{2n+1} = 2n^2+n,\quad n\in\mathbb{N}^+.
$$
In our case, set $n = 3$. Then,
$$\begin{align}S &= 2(3)^2 + 3 + 3 \\&=24 \end{align}$$
as desired.
