Which topic does it belong??

We construct a circumcircle. Thus, $2n\theta = 2\pi$ . And $\theta = \frac { \pi }{ n }$ .

$A_{1}A_{2} = 2R \sin { \theta }$ , $A_{1}A_{3} = 2R \sin {2 \theta }$ , $A_{1}A_{4} = 2R \sin {3 \theta }$ .

$\frac { 1 }{ \sin { \theta } } =\frac { 1 }{ \sin { 2\theta } } +\frac { 1 }{ \sin { 3\theta } }$

$\frac { 1 }{ \sin { \theta } } -\frac { 1 }{ \sin { 3\theta } } =\frac { 1 }{ \sin { 2\theta } }$

$\frac { \sin { 3\theta } -\sin { \theta } }{ \sin { \theta } \sin { 3\theta } } =\frac { 1 }{ \sin { 2\theta } }$

$\frac { 2\cos { 2\theta } \sin { \theta } }{ \sin { \theta } \sin { 3\theta } } =\frac { 1 }{ \sin { 2\theta } }$

$2\cos { 2\theta } \sin { 2\theta } =\sin { 3\theta }$

$\sin { 4\theta } =\sin { 3\theta }$

$4\theta =\pi -3\theta$

$\theta = \frac { \pi }{ 7 }$

n = 7

Easy Problem! I solved It by using Complex Number's !

Trick is that our answer is independent of reference Side Length of Polygon , So Let the vertex as the n' th root's of unity ! where Origin is cir-cum centre of origin

So Vertex is $\displaystyle{{ A }_{ k }\equiv ({ e }^{ i\cfrac { 2(k-1)\pi }{ n } })}$

$\displaystyle{{ A }_{ 1 }{ A }_{ k }=\left| 1-{ e }^{ i\cfrac { 2(k-1)\pi }{ n } } \right| =\left| 1-\cos { \cfrac { 2(k-1)\pi }{ n } } -i\sin { \cfrac { 2(k-1)\pi }{ n } } \right| \\ { A }_{ 1 }{ A }_{ k }=2\sin { \cfrac { (k-1)\pi }{ n } } \left| \sin { \cfrac { (k-1)\pi }{ n } } -i\cos { \cfrac { (k-1)\pi }{ n } } \right| \\ { A }_{ 1 }{ A }_{ k }=2\sin { \cfrac { (k-1)\pi }{ n } } \\ \because \cfrac { 1 }{ { A }_{ 1 }{ A }_{ 2 } } =\cfrac { 1 }{ { A }_{ 1 }{ A }_{ 3 } } +\cfrac { 1 }{ { A }_{ 1 }{ A }_{ 4 } } \\ \cfrac { 1 }{ \sin { \cfrac { \pi }{ n } } } =\cfrac { 1 }{ \sin { \cfrac { 2\pi }{ n } } } +\cfrac { 1 }{ \sin { \cfrac { 3\pi }{ n } } } \quad (Say\quad \cfrac { 2\pi }{ n } =\theta \quad )\\ \cfrac { 1 }{ \sin { \cfrac { \theta }{ 2 } } } \quad -\quad \cfrac { 1 }{ \sin { \cfrac { 3\theta }{ 2 } } } =\cfrac { 1 }{ \sin { \theta } } \\ \Rightarrow \sin { 2\theta } =\sin { \cfrac { 3\theta }{ 2 } } \\ 2\theta +\cfrac { 3\theta }{ 2 } =\pi \\ \theta =\cfrac { 2\pi }{ 7 } =\cfrac { 2\pi }{ n } \\ \boxed { n=7 } }$

Q.E.D

$\text{Note the relation of all angles to } \alpha~ and~ various ~isosceles~ triangles.$ $\text{Let n be the number of sides. } 2\alpha=\dfrac{360}{n}~\therefore~\alpha=\dfrac{360}{2n}.....(1)~.\\\text{The figure shows various angles formed between three sides of a n-gon.}\\ let~a=A_1A_2,~~b=A_1A_3,~~c=A_1A_4.~~Applying ~Sin~ Law~to~\triangle A_1 A_3 A_4,\\\dfrac{Sin\alpha}{a}=\dfrac{Sin2\alpha}{b}=\dfrac{Sin3\alpha}{c}~~\because Sin(180-3\alpha)=Sin(3\alpha)\\ \dfrac { 1 }{ { A }_{ 1 }{ A }_{ 2 } } =\dfrac { 1 }{ { A }_{ 1 }{ A }_{ 3 } } +\dfrac { 1 }{ { A }_{ 1 }{ A }_{ 4 } } \implies~\dfrac { 1 }{ a } =\dfrac { 1 }{b } +\dfrac { 1 }{ c }\\\dfrac{Sin\alpha}{a}=\dfrac{Sin2\alpha}{b}=\dfrac{Sin3\alpha}{c} \implies~1=\dfrac{1}{Sin\alpha}*\left(\dfrac{1}{Sin2\alpha}+\dfrac{1}{Sin3\alpha} \right) ..(2) \\From~ (1) \alpha= \dfrac{360}{2n}~. Below~ we~ see~ ~\alpha~for~diffrent~ n.:-\\n=5\rightarrow~\alpha~=36....~~~~~~~~~~~~~~~~~ n=6\rightarrow~\alpha~=30...\\\color{#3D99F6}{n=7\rightarrow~\alpha~=25.71}....~~~~~~~~~~~~~~~~~n=8\rightarrow~\alpha~=22.5.... \\From~~(2)~~Letf(\alpha)= \dfrac{1}{Sin\alpha}*\left(\dfrac{1}{Sin2\alpha}+\dfrac{1}{Sin3\alpha} \right) \\\text{Solve this fuction through calculator, graph in graphing calculator }\\\text{or otherwise for values given above so that } f(\alpha)=1. ~That~gives~n. \\ \color{#3D99F6}{f(25.71)=1. ~~~So~~~\Large n=7}.~~~~~~~As ~n~~increases~so~ does ~f(\alpha). \\$
After getting (2) you may follow Lavisha Parab 's method, or use calculator.
Lavisha Parab 's solution is a better one, but I have given this to show a slightly different approach

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after having 2 amazing solutions from my friends i have 1 more method..... $\angle(A1A2A3)=\frac{ (n-2)\pi }{ n }$ $A1A3=A2A4=2a \cos \frac{ \pi }{ n }$ $now,A1A2*A3A4+A2A3*A1A4=A1A3*A2A4$ $a^2+A2A3*A1A4=(2a \cos \frac{ \pi }{ n })^2$ $A1A4=a(4\cos^2\frac{ \pi }{ n }-1)$ $now, \frac{ 1 }{ A1A2 }=\frac{ 1 }{ A1A3 }+ \frac{ 1 }{ A1A4 }$ $2 \cos \frac{ \pi }{ n }[4\cos^2\frac{ \pi }{ n }-1]=4\cos^2\frac{ \pi }{ n }-1+2\cos \frac{ \pi }{ n }$ $\cos \frac{ \pi }{ n }=t$ $8t^3-4t^2-4t+1=0$ $answer->\cos \frac{ \pi }{ 7 },\cos \frac{ 3\pi }{ 7 },\cos \frac{ 5\pi }{ 7 }$ $n=7$