$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?

Hint:
Let, each side of the regular polygon $A_1A_2A_3\ldots A_n$ be $a$ then we have $$A_1A_2=a$$
$$A_1A_3=2a\sin\left(\frac{(n-2)\pi}{2n}\right)$$
$$A_1A_4=a-2a\cos\left(\frac{(n-2)\pi}{n}\right)$$ 
Now, we have 
$$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$
$$\frac{1}{a}=\frac{1}{2a\sin\left(\frac{(n-2)\pi}{2n}\right)}+\frac{1}{a-2a\cos\left(\frac{(n-2)\pi}{n}\right)}$$
Let, $\frac{(n-2)\pi}{2n}=\alpha$ then we have 
$$1-\frac{1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$
$$\frac{1-2\cos2\alpha-1}{1-2\cos2\alpha}=\frac{1}{2\sin\alpha}$$
$$8\sin^3\alpha-4\sin^2\alpha-4\sin\alpha+1=0$$
I hope you can take it from here.
