If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$$\frac {1}{1 + x} = y$$
$$(1 + x)y = 1\tag{1}$$
Similarly,
$$(1 + y)z = 1\tag{2}$$
$$(1 + z)x = 1\tag{3}$$
Please help me to continue from here.

From
$\cos A=\tan B,
\cos B=\tan C,
\cos C=\tan A
$,
$\begin{array}\\
\cos A 
&= \sin B/\cos B\\
&= \sin B/\tan C\\
&= \sin B/(\sin C/\cos C)\\
&= \sin B \cos C/\sin C\\
&= \sin B (\sin A/\cos A)/\sin C\\
\end{array}
$
or
$\cos^2 A = \sin A\sin B/\sin C
$.
Similarly,
$\begin{array}\\
\cos B 
&= \sin C/\cos C\\
&= \sin C/\tan A\\
&= \sin C/(\sin A/\cos A)\\
&= \sin C \cos A/\sin A\\
&= \sin C (\sin B/\cos B)/\sin A\\
\end{array}
$
or
$\cos^2 B = \sin B\sin C/\sin A
$.
Also,
$\begin{array}\\
\cos C 
&= \sin A/\cos A\\
&= \sin A/\tan B\\
&= \sin A/(\sin B/\cos B)\\
&= \sin A \cos B/\sin B\\
&= \sin A (\sin C/\cos C)/\sin B\\
\end{array}
$
or
$\cos^2 C = \sin A\sin C/\sin B
$.
Letting
$x=\sin A,
y = \sin B,
z = \sin C
$,
these become
$1-x^2 = xy/z,
1-y^2 = yz/x,
1-z^2 = zx/y
$.
Dividing the first two,
$\frac{1-x^2}{1-y^2}
=\frac{x^2}{z^2}
$.
From the last one,
$x = \frac{y(1-z^2)}{z}
$
so
$1-y^2
=\frac{zy}{\frac{y(1-z^2)}{z}}
=\frac{z^2}{1-z^2}
$
or
$y^2
=1-\frac{z^2}{1-z^2}
=\frac{1-2z^2}{1-z^2}
$.
Similarly,
$\frac{1-x^2}{\frac{z^2}{1-z^2}}
=\frac{x^2}{z^2}
$
or
$(1-x^2)(1-z^2)
=x^2
$
or
$1-z^2
=\frac{x^2}{1-x^2}
$
or
$z^2
= 1-\frac{x^2}{1-x^2}
= \frac{1-2x^2}{1-x^2}
$.
Similarly,
again,
we get
$\begin{array}\\
x^2
&= \frac{1-2y^2}{1-y^2}\\
&= \frac{1-2\frac{1-2z^2}{1-z^2}}{1-\frac{1-2z^2}{1-z^2}}\\
&= \frac{1-z^2-2(1-2z^2)}{1-z^2-(1-2z^2)}\\
&= \frac{3z^2-1}{z^2}\\
&= \frac{3\frac{1-2x^2}{1-x^2}-1}{\frac{1-2x^2}{1-x^2}}\\
&= \frac{3(1-2x^2)-(1-x^2)}{1-2x^2}\\
&= \frac{2-5x^2}{1-2x^2}\\
\text{or}\\
x^2-2x^4
&=2-5x^2\\
\text{or}\\
0
&=2x^4-6x^2+2\\
\text{or}\\
0
&=x^4-3x^2+1\\
\text{or}\\
x^2
&=\dfrac{3\pm\sqrt{9-4}}{2}\\
&=\dfrac{3\pm\sqrt{5}}{2}\\
\end{array}
$
Since
$0 \le x^2 \le 1$,
we must have
$x^2
=\dfrac{3-\sqrt{5}}{2}
$,
so
$x
=\pm\dfrac{\sqrt{5}-1}{2}
$.
Going through the same thing
for $y$ and $z$,
we get
$x = y = z
=\pm\dfrac{\sqrt{5}-1}{2}
$.
To determine the signs,
multiply
$1-x^2 = xy/z,
1-y^2 = yz/x,
1-z^2 = zx/y
$
together.
We get
$(1-x^2)(1-y^2)(1-z^2)
=xyz
$.
Therefore
$xyz > 0$
so that
0 or 2 of them
are negative
and the other(s)
are positive.
