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.
 A: $$\tan^2C=\cos^2B=\dfrac1{1+\tan^2B}=\dfrac1{1+\cos^2A}=\dfrac1{2-\sin^2A}$$
$$\cos^2C=\tan^2A$$
$$\implies\tan^2C=\sec^2C-1=\dfrac{1-\cos^2C}{\cos^2C}=\dfrac{1-\tan^2A}{\tan^2A}=\dfrac{1-2\sin^2A}{\sin^2A}$$
Equating the values of $\tan^2C$ and writing $\sin^2A=x$
$$\dfrac1{2-x}=\dfrac{1-2x}x\implies x^2-3x+1=0\implies x=\dfrac{3-\sqrt5}2$$
As $0\le x\le1$
A: Define $a := \cos^2 A$, and $b := \cos^2 B$, and $c := \cos^2 C$. Then 
$$\cos A = \tan B \quad\to\quad \cos^2 A = \tan^2 B = \sec^2 B - 1 \quad\to\quad a = \frac{1}{b} - 1 \quad\to\quad b = \frac{1}{1+a}$$
Likewise,
$$c = \frac{1}{1+b} \qquad\text{and}\qquad a = \frac{1}{1+c}$$
so that, adding a slightly-gratuitous $1$, 
$$1+a \;=\; 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a}}} \\ \tag{$\star$}$$
The evident recursion reveals that $1+a$ (and also $1+b$ and $1+c$) is the "$1$s all the way down" continued fraction, which some will recognize as representing the Golden Ratio, $\phi := \frac{1}{2}(1+\sqrt{5}) = 1.618\ldots$. Consequently,
$$a = b = c = \phi - 1 = \phi^{-1}$$
We finally calculate
$$\sin^2 A = \sin^2 B = \sin^2 C = 1 - \phi^{-1} = \phi^{-2}$$
so that 

$$|\sin A| = |\sin B| = |\sin C| = \phi^{-1} = 0.618\ldots$$

That the signs of the sines match is left as an exercise to the reader.
Observe that we'd reach the same conclusion no matter the length of the  "loop" of equations. Once $1+a$ appears, at any stage, in the right-hand-side of the counterpart of $(\star)$, the implied continued fraction collapses to the simpler form 
$$1 + a = 1 + \frac{1}{1+a} \quad\text{, giving}\quad a = \frac{1}{1+a}$$
which represents the one-equation loop, $\cos A = \tan A$.
A: If no two of $x,y,z$ are equal then all of them are unequal. Assume $x>y>z.$ From $(1) and (2)$ $y - z + y(x - z) = 0 $. This implies $ y < 0 $(since $y - z > 0, x - z > 0)$.....$(4)$. From $(2) and (3)$:$ z -x + z(y - x) = 0 $
This implies $z < 0$ (since $z - x < 0, y - x < 0)$ $......(5)$. 
From $(3) and (1)$: $x - y + x(z - y) = 0 $
This implies $x > 0$(since $x - y > 0, z - y < 0$)$.........(6)$ 
(4) and (6) contradicts (1), which has (1 + x)y = 1 
→ our assumption that $x, y, z$ are all unequal is incorrect and 
some two of them are equal. If $x = y$, $(2)$ becomes 
$(1 + x)z = 1 = (1 + z)x$ (by $(3)$) 
→$1 - z = xz = 1 - x → z = x→ x = y = z$. 
If $y = z$, it similarly follows that $x = y = z$. 
So we have: $tan²A = tan²B = tan²C$ 
→$A = B = C →sinA = sinB = sinC$.
A: 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.
