Question

If cos x = tan y, cos y = tan z, cos z = tan x; prove that -
sin x = sin y = sin z = 2 sin ${18}^{\circ }$

Answer 1

Making use of given relations , we have
$co{s}^{2}x=ta{n}^{2}y=se{c}^{2}y-1=co{t}^{2}z-1$
or $1+co{s}^{2}x=\frac{co{s}^{2}z}{1-co{s}^{2}z}=\frac{ta{n}^{2}x}{1-ta{n}^{2}x}$
$or1+co{s}^{2}x=\frac{si{n}^{2}x}{co{s}^{2}x-si{n}^{2}x}$
On changing to sin x, we get,
$(2-si{n}^{2}x)(1-2-si{n}^{2}x)=si{n}^{2}x$
$or2si{n}^{4}x-6si{n}^{2}x+2=0$
$orsi{n}^{4}x-3si{n}^{2}x+1=0$
$\therefore si{n}^{2}x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}=\frac{3-\sqrt{5}}{2}$
We have reject the value $\frac{3+\sqrt{5}}{2}$ as it is $>1$
or $si{n}^{2}x=\frac{6-2\sqrt{5}}{4}={\left(\frac{\sqrt{5}-1}{2}\right)}^2$
$\therefore sinx=\frac{\sqrt{5}-1}{2}=2.\frac{\sqrt{5}-1}{4}=2=sin{18}^{\circ }$
By Symmetry we can say that,
sin x = sin y = sin z = $2sin{18}^{\circ }$