If cosx =tany, cosy =tan z & cosz =tanx prove that sinx =siny =sinz

We have : cos x =tan y.
$\therefore co{s}^{2}x=ta{n}^{2}y=se{c}^{2}y-1\left(1\right)$
But $cosy=tanz\therefore secy=cotz$
$\therefore$ from (1),
$co{s}^{2}x=co{t}^{2}z-1$
$\therefore 1+co{s}^{2}x=co{t}^{2}z=\frac{co{s}^{2}z}{si{n}^{2}z}=\frac{co{s}^{2}z}{(1-co{s}^{2}z)}$
But cos z =tan x
$\therefore 1+co{s}^{2}x=\frac{ta{n}^{2}x}{(1-ta{n}^{2}x)}\therefore 1+(1-si{n}^{2}x)=\frac{\left(\frac{si{n}^{2}x}{co{s}^{2}x}\right)}{[1-(\frac{si{n}^{2}x}{co{s}^{2}x}\left)\right]}\therefore 2-si{n}^{2}x=\frac{si{n}^{2}x}{(co{s}^{2}x-si{n}^{2}x)}\therefore 2-si{n}^{2}x=\frac{si{n}^{2}x}{(1-2si{n}^{2}x)}$
$\therefore (2-si{n}^{2}x)(1-2si{n}^{2}x)=si{n}^{2}x\therefore 2si{n}^{4}x-6si{n}^{2}x+2=0$
$\therefore$ by Quadratic Formula,
$si{n}^x=\frac{[3\pm \sqrt{9-4}]}{2}=\frac{(3\pm \sqrt{5})}{2}$
But $\frac{(3+\sqrt{5})}{2>1}whereassi{n}^{2}x\le 1$
$\therefore si{n}^{2}x=\frac{(3-\sqrt{5})}{2}=\frac{(6-2\sqrt{5})}{4}=\frac{(\sqrt{5}-1{)}^2}{4}\therefore sinx=\frac{(\sqrt{5}-1)}{2}=2sin{18}^{\circ }$
We can similarly show that
$..sinx=siny=sinz=2sin{18}^{\circ }=\frac{(\sqrt{5}-1)}{2}$